Calc Problem?

Question

A rectangular box must have a volume of 2 cubic meters. The material for the base and top costs $ 2 per square meter. The material for the vertical sides costs $ 64 per square meter.


(a) Express the total cost of the box in terms of the length (l) and width (w) of the base.


C = $


(b) Find the dimensions of the box that costs least.
length = meters
width = meters
height = meters

Answer 1

Proceeding from the line C = 4LW + 128WH + 128LH in the above answer and using W = 2/LH we get C = 8/H + 256/L + 128LH. You now do partial differentiation on this. (I will have to use dC/dL and dC/dH as I can't write the proper partial derivative signs.) dC/dH = 128L - 8/H and dC/dL = 128H - 256/L^2. Putting each of these equal to zero gives HL^2 = 2 and LH^2 = 1/16. Combining these gives L = 4 and H = 1/8. Replacing these in W = 2/LH gives W = 4. It is no real surprise that the length and width are the same as the costing gave no preference to each. On the other hand the height is so small because the cost of side material was so high.

I've just realised that I've done the whole thing with C in terms of L and H rather than L and W as in your question. However that should go in a very similar manner.

Answer 2

(a) We're given that the base and top cost $2/square meter.
We're also given that the sides cost $64/square meter.

C = (cost of base and top) + (cost of 4 vertical sides)

The cost is represented by multiplying the cost per square meter times the surface area.
If we have L = length, W = width, H = height,

The area of the top and bottom would be LW.
The area of left and right would be WH.
The area of front and back would be LH.

C = 2(LW) + 2(LW) + 64(WH) + 64(WH) + 64(LH) + 64(LH)

Grouping like terms,

C = 4LW + 128WH + 128LH

We also know that V = LWH, so
2 = LWH

(b)

I just realized this involves a higher level of Calculus than I know (since more than 2 variables are involved, even with the volume condition).