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.*

======================================

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.*