Using the feasible region below, determine which x and y values will maximize the profit, P, in the equation P = 40x + 55y * 1 point A(0, 500)
B(300, 400)
C(380, 200)
D(400, 0)
E None of the Above

B (300, 400)

Step-by-step explanation:

The profit maximization will be when the sum of the products will be greater. The maximum profit will be when x is 300 and y is 400. If we put in the equation :

P = 40x + 55 y

A - When x =  0 , y = 500

P = [40 * 0] + [55 * 500]

P = 27500

B -

When x =  300 , y = 400

P = [40 * 300] + [55 * 400]

P = 34000

C -

When x =  380 , y = 200

P = [40 * 380] + [55 * 200]

P = 26200

D -

When x =  400 , y = 0

P = [40 * 400] + [55 * 400]

P = 16000

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w