Question

Help with calculus optimization problem?

A rectangular building will have 6250 square feet of floor space. It will be divided into 4 separate rectangular rooms by 3 parallel interior walls. The exterior walls will cost $110 per linear foot. The interior walls will cost only $75 per linear foot. Find the dimensions that will minimize the cost of the building.

Answer 1

Please see below.

Explanation:

We'll make the lengths of the exterior `x` by `y` and we'll make the interior walls parallel to the wall of length `y`, so they also have length `y`

To get area `6250`, we'll need `xy=6250`

We need two exterior walls of length `x` and two more of length `y`.
The cost of the exterior is `$110(2x+2y)`.

The cost of the three interior walls is `$75(3y)`.

The total cost is `C = 110(2x+2y) + 75(3y)`.

Using `xy=6250`, we get `x = 6250/y`, so we seek to minimize

`f(y) = 110(2(6250/y)+2y)+75(3y)`

`f(y) = (220(6250))/y+220y + 225y`

`f(y) = (220(6250))/y+445y`

Now we can find `y` to minimize `f` (the cost).

`f'(y) = (-220(6250))/y^2+445`

`f'(y) = 0` if and only if `220(6250) = 445 y^2`

`220(5(1250)) = 5(89) y^2`

`y^2 = (220(1250))/89`

`y = sqrt(22 * 10 * 10 * 25 * 5)/sqrt89`

`= 50 sqrt(110/89)`

Note that `f''` is negative for positive `y`, so `f(50 sqrt(110/89))` is a minimum.

And `x = (6250sqrt89)/(50sqrt(110)) = (125sqrt89)/sqrt110`

Get decimal approximations if needed and write the answer clearly.