Question

Please please help me out.........................

Answer 1

The constraints can be rewritten with the following subdivision along the x axis:

Case 1: 0 <= x <= 6

In this case, the constraint for y becomes

`y \geq -2x+20`

Which implies that

`x+y \geq x-2x+20=20-x`

The mimum for this function, given that x ranges from 0 to 6, is clearly 20-6=14.

Case 1: 6 <= x <= 18

In this case, the constraint for y becomes

`y \geq -(2)/(3)x+12`

Which implies that

`x+y\geq x-(2)/(3)x+12=12+(1)/(3)x`

The mimum for this function, given that x ranges from 6 to 18, is clearly

`12+(1)/(3)\cdot 6 = 12+2=14`

Case 1: x > 18

In this last case, any (positive) value for y is fine. Since x is at least 18, the sum of x and y will also be at least 18 (remember that y is non-negative).

So, out of all three cases, the minimum is achieved at x=6, y=8

Answer 2

C = 14 is the minimum value

Explanation:

Sketch the inequalities using

2x + y = 20

with x- intercept = (10, 0) and y- intercept = (0, 20)

2x + 3y = 36

with x- intercept = (18, 0) and y- intercept = (0, 12)

Solve 2x + y = 20 and 2x + 3y = 36 simultaneously to find

The point of intersection at (6, 8))

The vertices of the feasible region are at

(0, 20), (6, 8), (18, 0)

Evaluate the objective function C = x + y at each of the vertices

(0, 20) → C = 0 + 20 = 20

(6, 8) → C = 6 + 8 = 14 ← minimum value

(18, 0) → C = 18 + 0 = 18

Minimum value is C = 14 when x = 6 and y = 8