Answer :
x = 58.62 feet
y = 85.3 feet
Given that,
The cost of exterior walls is $140 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)(110)
For the interior wall, we have 100x
The cost function = C
⇒ C = 2(x + y)(110) + 100x
⇒ C = 220(x + y) + 100x
= 220x + 240y + 100x
= 320x + 240y
Recall that y = 5000/x
C = 320x + 220(5000/x)
C = 320x + 1100000/x
Differentiate C with respect to x
C'(x) = 320 - 1100000/x²
= (320x² -1100000) / x²
To minimize cost
C'(x) = 0
(320x² -1100000) / x² = 0
320x² -1100000 = 0
⇒ 320x² = 1100000
⇒ x² = 1100000/320
⇒ x = √1100000/320
⇒ x = 58.62 feet
Recall that y = 5000/x
y = 5000/58.62
y = 85.3 feet
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To find the dimensions that minimize the cost of building the warehouse, set up an equation based on the area of the two rooms and solve for the dimensions.
Dimensions of the warehouse:
- Let one side of the interior wall be x feet, so the other side will be 100 - x feet.
- Express the area of the two rooms in terms of x: x(100) = 5000 - x(100).
- Solve for x to find the dimensions that minimize cost.
