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I welcome back to my class in
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mathematics in the modern world for
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today we will take another example of a
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linear program maximization linear
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program but this time our linear program
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will consist of has four explicit
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constraints but the same two decision
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variables and we will solve this linear
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program by graphical method and we will
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draw the graphs in the first quadrant of
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the Cartesian plane because of the
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restrictions imposed by the implicit
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constraints that the decision variables
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should be positive or 0 for those of you
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who missed the lessons before in
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mathematics in the modern world search
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for my videos are teachers mathematics
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in the modern world our linear program
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is as follows maximize P equals 7x plus
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5y our decision variables are X and Y
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and we want to find these values of the
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decision variables x and y that will
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give the largest value of P subject
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through the following constraints the
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first constraint is a linear inequality
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that contains the less than equal to the
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less than equal to inequality 5x plus 8y
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less than equal to 14 this is the first
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explicit constraint and we denote it by
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1 the second constraint is another
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inequality that contains the less than
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equal do in the quality side 9x + 6 y
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less than equal to 54 this is the second
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explicit constraint and we benefit by 2
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the third constraint has only one
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decision variable X but still it is a
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linear inequality constraint that
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contains the greater than equal to
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inequality sign X greater than equal to
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2 is the third explicit constraint and
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we denote it by three and the fourth
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cosplay is also a linear inequality
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constraint containing only one decision
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variable y and y should be less than
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equal to 4
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these the fourth explicit constraint and
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these constraints that require x and y
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to be greater than equal to 0 are the
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implicit constraints or non negativity
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constraints our in the graphical method
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we sketch the graphs of the explicit
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constraints and we have four explicit
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constraints we do not drop the implicit
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constraints and we go to graph the
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objective function and we confine our
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graph in the first quadrant of the
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partition plane where we write X in
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Horace on
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sis and when I apply in the vertical
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axis at the intersection of these two
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accesses the origin then opened by point
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zero again we can find the graph in the
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first quadrant because of the
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restrictions imposed by the implicit
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constraints that actually be greater
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than equal to 0 and Y should be greater
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than equal to 0 we start with the first
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explicit constraint the graph of the
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linear inequality consists of two parts
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the graph for the equal part which is a
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straight line and the graph for the less
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than part which is say region we start
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with an equal point of the first
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explicit constraint we simply initially
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removed the less than sign and we get
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the linear equation 5x plus 8y equals 4t
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again we find the intercepts of the line
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whose equation is given by this linear
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equation 5x plus 8y equals 4t the
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intercepts are the points on the x axis
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and the y axis where the knife will pass
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through the intercept the x-axis is a
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point we coordinate y equal to 0 thus to
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find the x intercept of the line with
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the given equation of them from the
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given equation of the line we set Y to 0
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when y equals 0 we get 5x
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eight y is zero equals 48 by 0 is 0 0
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plus 5x is 5x equals 40 and the sub 4x
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the 5% of X must be 1
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must we be my both sides of the equation
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by 5 and we get x equals 40 divided by 5
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is 8 so the language pass through in the
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x axis at the point x equals a and y
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equals 0 and we approximate this point
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in the x axis I will use a one unit
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intervals one two three you can eight
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units from zero if you are using a one
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unit interval one five six seven
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this is the point with coordinates eight
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serum and the line will pass through
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this point the line with equation 5x
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plus 8y equals 4t this point has
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coordinates eight serum next we find the
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y-intercept of the line or the point at
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which the line will pass through the y
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axis a point on the y-axis has abscissa
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or body of x equal to zero thus to find
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the y-intercept of the line you set X to
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0 in the equation of the line you said X
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to 0 when x is 0 we get 5 x axis 0 by x
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x0 plus 8y equals 40
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and we suffer why from this equation we
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have five by zero is zero plus 8y v by
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zero is zero zero plus 8y is 8y equals
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40 and when you are solving for the body
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of a body but it's quite patient must be
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one thus we divide both sides of this
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equation by a to make the patient of y
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equal to 1 n equal personal from the
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left side and we get y equals 5 thus the
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y-intercept the point on the y axis has
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coordinates x is 0 and Y is 5 again you
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plot this point or approximate this
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point on the y axis if you are using one
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unit inter bus in the y axis you can
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find units from the origin 1 2 3 4 5
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this point has coordinates 0 5 the line
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will pass through this point with
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coordinates 0 5 now you will connect
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these two intercepts with a straight
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line and you have the graph of the line
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whose equation is 5x plus 8y equals 14
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and that's the Y as
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we the x-intercept
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and we label the this language X
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equation 5x plus 8 1 by X plus 8y equals
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or this is the graph of the for the
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equal part of the first explicit code
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string for the less than part of for the
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less than part for the graph for the
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last part of these explicit constraints
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we shade shade the region below the line
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she rejoined below the line
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if you have greater done in the explicit
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constraints for strength with that these
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linear in two body once you shade the
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the region above the line if this is 5 x
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+ 8 y greater than equal to 40 for the
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greater than part if this is greater
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than 14 you shade the region above the
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line all the points in this region the
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region below the line consists of points
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whose values for x and y were
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substituted into 5 x + 8 Y will give you
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a sum of less than 40 and all the points
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on the line give you a combination of x
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and y values of x and y when you
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substitute - 5 x + 8 1 will give you a
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sum of exactly 40 so this the completed
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graph for the first explicit constraint
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we now sketch the graph of the second
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explicit constraint honestly the first
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explicit constraint the graph of the
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second explicit constraint has two parts
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the graph for the equal part which is a
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straight line the graph for the equation
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9x + 6 y equals 54 and the graph for the
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less than part which is a region we
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start with the graph of the equation 9x
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+ 6 y equals 54 now in X plus 6y
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it was 54 initially you removed
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inequality sign you will get a linear
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equation in two variables and we know
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that the graph of a linear equation in
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two variables is a straight line that
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has an x-intercept and y-intercept and
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when you find the x-intercept because
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the x-intercept is a point on the x-axis
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and any point on the x-axis has ordinate
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on serum thus when you find the
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x-intercept you simply set Y to 0 and
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substitute y equals 0 in the linear
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equation we said Y to 0 y equals 0 and
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we substitute 0 for y in this linear
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equation we have 9x plus 6 by 0 equals
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54 and we solve for X from the last
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equation we have 9 X 6 by 0 is 0 0 plus
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9 X is 9x equals 54 and the software
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asks me divide both sides of the
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equation by 9 now we are passing up from
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the left hand side and we get the value
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of x which is 54 divided by 9 equals 6
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and x-intercept the point at which the
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the graph of these linear equation will
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intersect the x-axis is the point with
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coordinates 6 and y 0 we plot this point
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on the x-axis by counting 6 units from
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the origin if you are using a 1 unit
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intervals in your horizontal axis 1 2 3
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4 5 6 so this point has coordinates 6
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theorem the the straight line with
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equation 9 X 6 y equals 54 we pass
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through this point with coordinates
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6 0 this is the x intercept of the line
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with this equation now we find the point
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on the vertical axis or the y axis where
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the language pass through and that point
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is called the y intercept of the line
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the y intercept is a point on the y axis
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and any point on the y axis the abscissa
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of any point are the body of X of any
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point on the y axis is serum has to find
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the y-intercept a point on the y axis
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you set X to 0 when x is 0 you get 9 x 0
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stick swine
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it was built before 1900 zero plus six Y
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is six one six y equals 54 and we make
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the pie efficient of y1 by dividing both
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sides up the last equation by 6 6 mil
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percent up from the left side and we get
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y equals before I divided by 6 is 9 and
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the point on the y axis at which the
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line will pass through has coordinates x
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is 0 and Y is 9 again you plot this
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point this is a point on the y axis
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because x is 0 and from 0 you count nine
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units upward if you are using a 1 unit
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inter bus so this 5 6 7 8 9 and the line
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the line we base equation 9x plus 6y
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equals 54 will pass through this point
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and coordinate coordinates of this point
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are 0 9 x is 0 and Y is name
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now because not the line by connecting
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the y intercept and x intercept of the
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line and we label this line with its
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equation its equation is 9x plus 6y
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equals 54 these nine has equation 9x 6y
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equals 54 that's the first part of the
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graph of the second explicit posturing
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for the equal part for the less than
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part again you shade the region below
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the line if you have breaker down in the
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given explicit cosplay you shade the
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region above the line if you only have
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an equation as specific constraint then
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there is nothing to shade below or above
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the line if the explicit constraint is
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just an equation the graph is just
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saying line and there is nothing to
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shame above or below the line for the
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less than part of the graph of the
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second X Pacifica strain we shade the
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region below this line with this
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equation saying the region below the
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line
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all the points in the region below this
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this line will give you a sum of less
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than 50 porn again all the points in the
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region will not the second line with
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this equation 9x plus 6y equals 54 will
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give you combinations of x and y bodies
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weights when we substitute in this
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expression will give you a sum of less
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than pig before and all the points on
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the line
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will give you a combinations of the
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bodies of x and y which when you
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substitute into this expression will
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give you a sum of exactly equal to pig
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before
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now you sketch the graph of the third
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explicit constrain it's a linear
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inequality and it's linear in only one
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variable X the graph of x greater than
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equal to two has also two parts the line
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for the equal part or the line for the
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equation x equals two
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and the region for valleys of X greater
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than P or this linear equation in only
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one Bible as the brow is a vertical line
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this vertical line if you have a linear
00:22:43
equation in only one body of all X the
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graph is a vertical line and the
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vertical line is parallel with the
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y-axis and this line will pass through
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the point in the x axis x equals two so
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will that we locate x equals two in the
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x axis this is one and this point has
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coordinates X is two and Y is zero the
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graph of a linear equation x equals two
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will pass through this point
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and the line is vertical parallel with
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the y axis so construct that line
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through this point two zero vertical
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line parallel with the y axis I will
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construct that line this is the line x
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equals 2 all the points of this vertical
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line has abscissa of food but the
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ordinance of the point steeper only the
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abscissa of the points in on this
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vertical line are equal to or the
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greater that part we want the region
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this is our reference the line x equals
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2 and you want the region where the
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points where the abscissas of the points
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are greater than 2 we want the region
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where the bodies of X are greater than 2
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and this is 2 the numbers 2 to the lap
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up two are less than T and the numbers
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to the right of 2 are greater than 2 has
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4 x greater than 2 we shade the region
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to the right of x equals so stating that
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region or x greater than 2 all the
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points to the right of the line x equals
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2 the bodies of X are greater than 2 all
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the points to the right of x equals to
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the value of x values of x are greater
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than so this the complete and round of
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the third explicit per string and the
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last x Pacific restrain the port at
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specific restraint is also a linear
00:25:44
inequality in only one variable Y and
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the graph of this specific
00:25:52
because all set to parts the graph for
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the equal part or the graph for the
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equation y equals porn and the graph for
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y less than 4 which is same region so
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you notice if you have a linear equation
00:26:15
in one or two party monster graph is a
00:26:20
line but if you have a linear inequality
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in two or one body about the graph of
00:26:30
the linear inequality is a region or an
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area in the partition plane or y equals
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four the graph is saying horizontal line
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if you have a linear equation in only
00:26:50
one variable Y the graph is horizontal
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line and this horizontal line is
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parallel with the x-axis for this
00:27:01
particular linear equation y equals four
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the horizontal line will pass through
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the y-axis y equals four so you locate
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one equals four in the vertical axis or
00:27:15
y axis and this is the point y equals
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four this point has coordinates x is 0
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and Y is born
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the horizontal line with equation y
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equals four will pass through this point
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we intersect the y axis at this point
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and Horizonte line is parallel with the
00:27:43
x axis so we now passed up that line we
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draw the line
00:27:59
this is the line y equals porn for the
00:28:09
the graph for the less than part of Miss
00:28:13
explicit constrain for the graph of y
00:28:15
less than four you shrink the region
00:28:19
below the below the line y equals four
00:28:23
because all the points below the line y
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equals four the police of wine are less
00:28:30
than four if you have y greater than
00:28:34
form then you shade the region above the
00:28:38
line y equals four but in the fourth
00:28:42
expressive constraint we are given that
00:28:44
y should be less than four and the
00:28:50
points weight y or they need wine less
00:28:53
than four are the points below the line
00:28:56
y equals four this is our reference
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point the numbers below this point are
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less than four as we shake the region
00:29:06
below the line y equals four you know
00:29:12
the line y equals four and that's the
00:29:20
complete that's the complete graphs for
00:29:25
all the or explicit constraints the next
00:29:32
step is we will identify the feasible
00:29:35
solution area or feasible region the set
00:29:39
of points that will satisfy all the
00:29:43
explicit constraints and those set of
00:29:48
points that satisfy all the explicit
00:29:51
constraints is also a region and there's
00:29:57
only a polygon we will now identify the
00:30:06
feasible solution
00:30:08
area or the region consisting of all
00:30:11
points that satisfy all the constraints
00:30:13
both the expressive constraints and
00:30:18
implicit Gosling's again this is the
00:30:23
complete graph of the four explicit
00:30:27
constraints for 9 x + 6 y less than
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equal to 54 the proud possessor of this
00:30:36
straight line may be basic equation 9x
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plus 6y equals 54 and the region below
00:30:45
the line for the second explicit
00:30:49
constraint by x + 8 y less than equal to
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40 the graph essence of this straight
00:30:57
line with equation 5x plus 8y equals 40
00:31:02
and the region below this line for the
00:31:09
third explicit constraint x equals to or
00:31:16
greater than 2x greater than equal to 2
00:31:18
the purposes of this vertical line
00:31:21
parallel we know y axis and passing
00:31:24
through the point x equals 2 on the x
00:31:27
axis and the region to the right x
00:31:32
equals 2 where the bodies of X are
00:31:35
greater than 2 and the last explicit
00:31:38
customi y less than equal to 4 / c sub
00:31:41
base horizontal line parallel with the x
00:31:44
axis and passing through the through
00:31:47
this point 0 4 or why it was 4 on the y
00:31:51
axis and region below the line where the
00:31:56
beliefs of wine are less than for the
00:32:02
peaceable solution area will consists of
00:32:05
all points that satisfy all these for a
00:32:09
specific
00:32:10
straights the implicit constraints are
00:32:13
already satisfied by all the points in
00:32:17
the first quadrant and from the graph
00:32:20
the set of points that satisfy all the
00:32:23
mistakes are the points in a region
00:32:27
where you have four intersecting lines
00:32:32
you have to graph your you have to make
00:32:35
your graph very clear so you can easily
00:32:37
identify the region we for intersecting
00:32:41
lines in my graph that region we four
00:32:46
intersecting lines is a polygon bounded
00:32:50
by this part of the horizontal axis this
00:32:57
part of the positive x-axis this part of
00:33:02
the vertical line x equals to this part
00:33:10
of the horizontal line y equals port
00:33:17
it's also bounded by this part of this
00:33:22
this is great line with equation 5x plus
00:33:25
8y was 14 and this part this part this
00:33:35
part out this line with equation 9x plus
00:33:39
6y it must be before this is the
00:33:43
feasible solution area and we identify
00:33:48
like this by the acronym f as a or
00:33:54
feasible solution area principal
00:33:59
solution area
00:34:04
on the point saying these polygons with
00:34:09
sex 1 2 3 4 5 is we satisfy all the four
00:34:18
constraints again all the points in this
00:34:22
feasible solution area which is a
00:34:26
polygon with 1 2 3 4 5 6 will satisfy
00:34:32
all the four explicit constraints and
00:34:36
one of them will give us the values for
00:34:40
the decision variables that will give
00:34:43
that will produce the maximum value of B
00:34:48
now to because this feasible solution
00:34:53
era consists of infinitely many possible
00:34:56
solutions to find the values of x and y
00:35:01
that we the maximum value of P we limit
00:35:05
the evaluation of the objective function
00:35:08
to all those corner points thus peaceful
00:35:14
solution amia or we evaluate the
00:35:17
objective function the next step now is
00:35:20
to evaluate the objective function of
00:35:22
the vertices or corner points of the
00:35:27
feasible solution area and these are the
00:35:32
corner points or vertices of the
00:35:38
feasible solution area
00:35:44
we have to evaluate the objective
00:35:47
function the point of our ERP are the
00:35:53
bodies of X and mine each of these five
00:36:00
corner points or vertices of the
00:36:03
feasible solution area one of these
00:36:06
points will give us the maximum value of
00:36:10
P
00:36:22
[Music]
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[Music]
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[Music]
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is
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[Music]
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easy
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[Music]
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Oh
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I gotta find the words
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[Music]
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[Music]
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[Music]
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[Music]
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[Music]
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in
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[Music]
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and I do
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[Music]
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this
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[Music]
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[Music]
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we construct a stable for evaluating the
00:39:43
objective function our objective is to
00:39:46
find the maximum value of P which is
00:39:51
equal to 7x plus 5y and we evaluate the
00:39:56
objective function of each of these five
00:40:00
corner points of our peaceable solution
00:40:03
area we start with the corner point
00:40:06
whose coordinates are known as we start
00:40:12
with this point this is same vertex our
00:40:14
corner point of this piece of a solution
00:40:17
area and the coordinates of this point
00:40:19
are to 0 this point to zero the body of
00:40:27
X is to the body of Y is zero and you
00:40:36
substitute x equals 2 and y equals 0
00:40:40
into this equation for P we have seven
00:40:45
access two plus five x wine serum again
00:40:54
7 by 2 is 14 plus PI by zero is zero as
00:41:01
P equals 14 plus zero is 14
00:41:09
was working the partner point to zero or
00:41:18
the vertex to zero the body of be the
00:41:21
body of access to the body of point zero
00:41:24
when substitute that into this equation
00:41:27
gives a value of P equals fourteen the
00:41:32
next vertex or corner point with known
00:41:35
coordinates is this point six zero we
00:41:42
like that in our table the point six
00:41:51
zero the body of X is 6 the value of y
00:41:56
is 0 and we substitute 6 for X and 0 for
00:42:07
y again 7 by 6 is 42 plus PI by y 0 we
00:42:19
get 7 multiplied by X is 6 7 by 6 is 42
00:42:26
plus 5 times 0 is 0 so B is 42 plus 0
00:42:33
equals 40
00:42:42
the next birth backs we considered the
00:42:49
vertex or corner point of our peaceable
00:42:53
solution area this point thinks the
00:42:57
intersection of these vertical line x
00:43:01
equals 2 and horizontal line with
00:43:06
equation y equals 4 has the coordinates
00:43:10
of this point of intersection this
00:43:14
vertical line and this horizontal line
00:43:20
are X is 2 and y equals 4 the
00:43:30
coordinates of this point in our 2 & 4
00:43:33
because means the intersection the point
00:43:36
of intersection of the vertical line x
00:43:39
equals 2 and the horizontal line y
00:43:42
equals 4 and we like that in the table
00:43:46
the corner point with coordinates 2 & 4
00:43:52
this point the value of x is 2 and the
00:43:57
value of y is 4 and we substitute these
00:44:03
values of x and y into this equation to
00:44:06
find the value of P we have 7 x x axis 2
00:44:12
plus pi x y y is 4 or 7 by 2 is 14 plus
00:44:22
5 by 4 is 20 therefore P equals 14 plus
00:44:30
12 P is 34
00:44:36
the body of our P is 34 when X is 2 and
00:44:44
Y is 4 or at the corner point 2 for the
00:44:50
point of intersection of the vertical
00:44:52
line with equation x equals 2 and r is
00:44:57
on the line with equation y equals 4 you
00:45:01
can see from this table that this corner
00:45:04
point gives the maximum value of P among
00:45:09
these three corner points but that is
00:45:12
not yet our optimal solution or best
00:45:16
solution or the largest value of P
00:45:18
because we still have two corner points
00:45:21
to consider one of which is this point
00:45:31
this point thinks the intersection of
00:45:34
the line y equals 4 y equals 4
00:45:40
and this line with equation 5x + 8 y
00:45:48
equals 45 X plus 8 y equals 14 and to
00:46:00
find our ordinates of this point of
00:46:02
intersection of this horizontal line and
00:46:05
this line with this equation we must
00:46:10
solve these two linear equations
00:46:12
simultaneously and this is quite easy to
00:46:17
solve because Y is already 4 we already
00:46:21
know the value of y Y X 4 we simply
00:46:25
substitute Y it was 4 into this linear
00:46:28
equation to solve for X and we do that 5
00:46:33
X + 8 x y is 4 equals 14
00:46:45
now you solve this linear equation in
00:46:49
one variable X transpose the constant
00:46:53
term to the other side of the equation 8
00:46:58
by 4 is 32 when you transpose possibly
00:47:02
32 to tonight it becomes negative 32 4
00:47:06
from the left hand side we have 5x
00:47:08
equals 40 minus 32 or 5x equals 8
00:47:23
and the software actually by both sides
00:47:27
of mass equation by 5 5 a percent up
00:47:34
from the from the left side and you get
00:47:39
x equals now we used 2 decimal places
00:47:46
for our answer for X 8 divided by 5 is 1
00:47:54
point 6
00:48:03
one point six is up to one decimal place
00:48:07
if you want your answer to two decimal
00:48:10
places you will simply add 0 after 6 so
00:48:17
this answer now one point 60 is in two
00:48:20
decimal places and the zero here becomes
00:48:24
significant because without the serum
00:48:27
here your answer will only be in one
00:48:29
decimal place has the point of
00:48:33
intersection of this line y equals four
00:48:37
and this night five x plus 8y equals 14
00:48:42
has coordinates x one point sixty one
00:48:48
point six p x when y equals four and we
00:48:55
write that in the k-12 evaluate the
00:49:01
objective function this corner point one
00:49:08
point 16 and for these this corner point
00:49:21
we this coordinates the value of x again
00:49:25
is one point sixty and the value of y is
00:49:34
4 for these values of x and y we about
00:49:41
way we find the value of p by replacing
00:49:46
absolutely one point sixty and replacing
00:49:50
y v4 and you may use calculator to
00:49:55
compute for the value of P when X is 1
00:49:59
point 16 and y equals four we now
00:50:05
substitute the value of x
00:50:09
1.60 x 7 + pi x y y is for again you may
00:50:19
use your calculator to compute for the
00:50:23
body of B we have seven times one point
00:50:29
sixty plus five x four and the sum is
00:50:46
thirty one point five
00:50:57
the corner point the coordinates of this
00:51:02
point are 1.6 P and 4 and when the
00:51:09
objective function is evaluated at this
00:51:12
corner point my P is thirty one point
00:51:17
five and among these corner points still
00:51:21
this corner point with coordinates
00:51:23
six-year-old gives the maximum body of P
00:51:27
but again that is not yet the best
00:51:31
solution or optimal solution because we
00:51:35
still have one corner point to consider
00:51:37
and you can see that this point is the
00:51:48
point of intersection of this line with
00:51:52
equation 9x plus 6y equals 54 and this
00:51:59
line find x plus 8y equals 40 and to
00:52:06
find the coordinates of this point of
00:52:09
intersection of these two lines we solve
00:52:13
the equations these two lines
00:52:17
simultaneously by using an algebraic
00:52:21
elimination method again we solve this
00:52:25
equation 9 x 9 x plus 6y equals 50 more
00:52:38
and the equation of this line 5x plus 8y
00:52:43
equals 14
00:52:49
I mean you said Ignatian by addition I
00:52:56
will eliminate variable X by elimination
00:53:02
by addition and by using elimination by
00:53:06
addition he should make the PI patients
00:53:08
of X numerically the same path with
00:53:11
opposite signs and to make me the five
00:53:16
patients of I've seen these two
00:53:18
equations numerically the same we
00:53:22
multiply this equation by the
00:53:26
coefficient of x in the second equation
00:53:28
find and we multiply the second equation
00:53:32
by the PI patient of apps in the press
00:53:37
equation 9 since we want to eliminate X
00:53:42
by subtraction one of the coefficients
00:53:46
of X must mean since we want to
00:53:50
eliminate X by addition one of the five
00:53:53
fishin's of X should be negative I will
00:53:57
make five mega P and we distribute
00:54:01
negative 5 to both sides of the first
00:54:04
equation we also multiply both sides of
00:54:08
the second equation by 9
00:54:18
negative five x a spade 9x
00:54:22
he's never be 45 X negative five x
00:54:28
positive six y is negative 30 wine 54 x
00:54:38
negative 5 is negative 270 now equation
00:54:48
both sides of this equation two x
00:54:50
Pacific 9 9 x 5 axis 45s 9 x positive 8
00:55:01
a 70 and y equals 9 x 40 is 360 as we
00:55:16
said we eliminate the variable X by
00:55:19
addition and you can now see that the
00:55:23
point patients of I've seen that the
00:55:25
equations are numerically the same both
00:55:27
are equal to 45 but they have opposite
00:55:31
signs one is negative and the other is
00:55:33
positive
00:55:34
thus we can eliminate X if we add these
00:55:39
two equations
00:55:41
negative 45 plus must be fortified will
00:55:44
add up to 0 and this is 72 72 y 9 x a AP
00:55:54
72 wine
00:56:01
and here we have negative 30 plus 72
00:56:08
he's possibly 42 possibly 42 y equals we
00:56:21
have 360 negative 270 plus 316 is
00:56:29
positive 90 finally we divide both sides
00:56:38
of the designation by poor people 19
00:56:44
divided by 40 to 42 Mia cancel out from
00:56:50
the left side of that mission
00:56:53
and we got a wine y equals 19 divided by
00:57:03
42 again used two decimal places when
00:57:08
you D by 90 by 42 we have 90 divided by
00:57:17
42 equals the two decimal places it's 2
00:57:24
point 1 4 y is 2 point 1 4 so the
00:57:36
ordinate of this point the point of
00:57:40
intersection of this line and this line
00:57:42
is 2 point 2 point 1 4
00:57:55
now we find out sisa of this point of
00:57:58
intersection and we do that by
00:58:01
substituting y equals 2 point 1 4 into
00:58:07
this equation into this equation or this
00:58:12
equation I will use the first equation
00:58:16
9x plus 6y equals 54 I will replace Y we
00:58:23
two point one four and I will solve for
00:58:26
X we have 9 x 9 x + 6 x wine our Y is 2
00:58:44
point 1 4 y is 2 point 1 4
00:58:56
it was before
00:59:04
and this equation is a linear equation
00:59:09
in the variable X and we solve for x
00:59:13
from this equation multiply six by two
00:59:16
point one four and response the product
00:59:19
to the right-hand side of the equation
00:59:21
and you only have 9x at the left side of
00:59:25
the equation 9x equals bit before you
00:59:33
transpose positive to the other side you
00:59:36
have negative six x 2.14 you multiply
00:59:45
six x to point one four and we get 12
00:59:51
point 84 we subtract subtract 12 point
00:59:59
84 prompt it before the Left we have 9x
01:00:03
equals 54 minus point 84 and we get
01:00:14
forty one point one six or one point one
01:00:20
six now divide both sides of the
01:00:24
numerical coefficient of x 9 make my
01:00:31
three cent of x equals 1 the point forty
01:00:38
one point was 6/9
01:00:44
goodbye by nine and we get 4.57 axes
01:00:55
4.57 so they'll seize up this point of
01:01:03
intersection of the two lines is 4.57
01:01:13
this corner point of the peaceable
01:01:17
solution area with these coordinates we
01:01:21
evaluate the objective function to find
01:01:23
the value of P when XX 4.57 and y is 2
01:01:30
point 1 4 find the value of our
01:01:36
objective function or the value of p for
01:01:39
the last corner point of our peaceable
01:01:43
solution area these corner points with
01:01:47
coordinates 4.57 and 2.1 for this corner
01:01:54
point the body of x is for point and 6
01:01:57
4.57 and the body of y is 2 point 1 4 in
01:02:03
our objective function with a place else
01:02:05
we'd 4.57 and substitute to point 1 4
01:02:11
for y we got seven x 4.57 plus 5 x 2
01:02:21
point 1 4 and finish our calculator to
01:02:24
check the the body of p7 x 4.57 plus 5 x
01:02:34
2 point 1 4 it was 4.16 and now by
01:02:43
inspection of this table can be seen
01:02:49
that the highest the maximum value of P
01:02:51
is 42 point 16 and and we get this
01:02:57
maximum value of P 1 F Series for 4.57
01:03:01
and y is 2 point 1 4 and this is our
01:03:06
best solution our optimal solution we
01:03:10
write x is 4.57
01:03:18
y is 2 point 1 4 and our maximum p value
01:03:37
the maximum value of P is 42 point 69
01:03:49
but all of these all of these are also
01:03:53
solutions except that the last corner
01:03:58
point give us gives us the best solution
01:04:01
or optimal solution this corner point
01:04:03
where X is one point 16 and Y is 4 keep
01:04:09
a value of P of thirty one point two is
01:04:13
also a solution this corner point with x
01:04:17
equals two and y equals four that a
01:04:21
value of P thirty-four is also a
01:04:24
solution
01:04:26
luckily all the points all the
01:04:31
infinitely many points in the piece of a
01:04:33
solution
01:04:33
amia our solutions but this is the
01:04:37
values of x and y that in the maximum
01:04:41
value of p has this is our best solution
01:04:45
x equals 4.57 and y equals two point one
01:04:50
four and the maximum value of P is forty
01:04:54
two point 69 for our next lesson we will
01:04:58
take an example of the minimization
01:05:02
problem that we were solved by graphical
01:05:06
method and if you have questions for
01:05:10
today's lesson you write your comments
01:05:14
in the you write your question in the
01:05:16
comments below but then I forget to
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01:05:22
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