Arectangle warehouse will have 5000 square feet of floor space and will be separated into two rectangular rooms by an interior wall. the cost of the exterior walls is $150 per linear foot and the cost of the interior wall is $100 per linear foot. find the dimensions that will minimize the cost of building the warehouse.

x = 59.4 ft

y = 84.18ft

Step-by-step explanation:

The cost of exterior walls is $150 per linear foot.

The cost of interior walls is $100 per linear foot.

xy = 5000

y = 5000/x

For the exterior walls, we have 2(x+y)(120)

For the interior wall, we have 100x

The cost function = C

C = 2(x+y)(120) + 100x

C= 240(x+y) + 100x

= 240x + 240y + 100x

= 340x + 240y

Recall that y = 5000/x

C = 340x + 240(5000/x)

C = 340x + 1200000/x

Differentiate C with respect to x

C'(x) = 340 - 1200000/x^2

= (340x^2 -1200000) / x^2

To minimize cost C'(x) = 0

(340x^2 -1200000) / x^2 = 0

340x^2 -1200000 = 0

340x^2 = 1200000

x^2 = 1200000/340

x = √1200000/340

x = 59.4 ft

Recall that y = 5000/x

y = 5000/59.4

y = 84.18 ft

step-by-step explanation:

most likely the answer is a.

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