# The feasible region region is the set of points on and inside the triangle with vertices (0,0),(8,0) and (0,10) find the maximum and minimum values of the objective function Q over the feasible region of the function a)Q=-4x-3y b)Q=10x-8y

1
by sweetysiri92

2015-04-02T01:34:55+05:30

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See diagram.
we have to find the value of Q at the vertices of the region and at the boundaries specified by the sides of the triangle. (the constraints).

The equations of the lines OA : y = 0.
OB :  x = 0.          and
AB :  x/8 + y/10  = 1      the intercept form.
5 x + 4 y = 40

The region shaded in the diagram = the area of the triangle = feasible region.
the region is specified by  x ≥ 0,  y ≥ 0,  5 x + 4 y ≤ 40
or,  y ≤ (10 - 1.25 x)

So the region is :  0 ≤ x ≤ 8,    0 ≤ y ≤ (10 - 1.25 x)

1)
Q = - 4 x - 3 y
≥ -4 x - 3 (10 - 1.25 x)
=>  Q  ≥  - 0.25 x - 30
The minimum value of Q : at x = 8,  is  -32
The maximum value of Q : at x = 0,  is x = - 30.
Value of Q at vertex O:  -4 * 0 - 3 * 0  = 0
at vertex A :  - 4 * 8 - 3 * 0 = - 32
at vertex B:  - 4 *0 - 3 * 10 = - 30

so  Q is maximum at the origin and is 0.  Q is minimum at A (8, 0) and is -32.
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2)  we have:   the region :  0 ≤ x ≤ 8,    0 ≤ y ≤ (10 - 1.25 x)

Q = 10 x - 8 y
≥ 10 x - 8 (10 - 1.25 x)
=>  Q ≥ 20 x - 80    or 20 (x - 4)
=> when x = 0, Q = -80, and when x = 8,  Q = 80.

Q is 0 at the origin O.
is  10 * 8 - 8 * 0 = 80 at  A (8, 0)
is   10 * 0 - 8 * 10 = - 80 at  B( 0, 10)

Thus the maximum value is at A (8,0) :  80
and the minimum value is at B (0, 10) :  -80.