A rectangular 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.

This is a calculus optimization problem. The aim is to minimize the total cost of building the warehouse. This involves forming our constraint and objective function, substituting the constraint into the objective, differentiating, and finding the value that minimizes cost.

Explanation:

This problem involves calculus and optimization. Our objective is to minimize the cost of building the warehouse which means we need to minimize the total length of walls. Another condition is that the warehouse has to have 5000 square feet of floor space.

Let's call one side of the rectangle "x" and the other side "y". The area (A) of the warehouse is given by A = xy = 5000. This equation is our constraint which expresses y in terms of x: y=5000/x.

The total cost (C) of building the warehouse is given by the sum of the cost of the interior and exterior walls. Our objective function becomes: C = 150(2x+2y) + 100y = 300x + 350y.

Substituting y from our constraint into the objective function we obtain an equation solely in terms of x, which we differentiate with respect to x, equate the derivative to zero, and solve for x to find the length that minimizes cost. The second derivative test verifies the minimum. The value of y can then be derived from the constraint.

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akivieobukomena akivieobukomena · Answer 2 · 2025-05-22T15:32:02-04:00

Answer:

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