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.
- 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
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.
- CaroleLv 44 years ago