# Optimization Problem- Differential Calculus?

A builder intends to construct a storage shed having a volume of 900 ft^3, a flat roof, and a rectangular base whose width is three-fourths the length. The cost per square foot of the materials is $4.00 for the floor, $6.00 for the sides, and $3.00 for the roof. What dimensions will minimize the cost?

Can you please tell why we use the following formula to compute the cost:

C= 4(wl) + 6(2lh) + 6(2lh) + 6(2wh) + 3(wl)

I'm confused why we should multiply these values to 4, 6, and 3. Please give a thorough explanation considering the above formula and explain each input in the formula such as 6(2lh) or 6(2wh). Thank you.

### 2 Answers

- cryptogramcornerLv 68 years agoFavorite Answer
The 4, 6, and 3 account for the cost per square foot of the materials.

For example, the area of the floor is width x length, which has been abbreviated wl

so the cost of the material of the floor, at $4 per square foot is 4(wl)

The shed has 4 sides. there are two with area length x height and 2 with areas width x height.

The cost of material is $6 per square foot, so you have

6(lh) and 6(2wh)

( you also seem to have an extra 6(2lh) in what you wrote; there should only be one

6(lh) term.)

The final term 3(wl) accounts for the roof, which costs $4 per square foot.

You also have a fixed volume to account for wlh = 900

Your next step will be to get rid of the w variable, replacing it with (3l/4) in all occurrences.

You volume equation would then be

(3l^2/4)h = 900 This can be solved for h

h = (4/3)900/(l^2) or

h = 1200/(l^2)

You would then use this to get rid of h in your cost equation. At this point

you will have C as a function of only l. You can take derivative, set to 0 and solve for

the optimal value of l. w will be (3/4) of whatever you get for l, and h can be found using

your h = 1200/(l^2) formula.

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- Anonymous4 years ago
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