A box of volume 72m^3 with square bottom and no top constructed out of two different materials. The cost of the bottom is $40/m^2 and the cost of the sides is $30/m^2. Find the dimensions of the box that minimize the total cost.

Question

How do I set up the equation(s) in the beginning and what is the derivative?

Answer 1

Let
h=height of box,
s=length=width of bottom

Constraint:
Volume, V = hs² = 72

Cost:
Cost, C = 40s²+4*30sh

Let λ=Lagrange multiplier
Objective function, F
F(s,h,λ)=40s²+4*30sh + λ(hs²-72)

Find partial derivatives with respect to each of the three variables and solve for s and l.

I get 3*4^(1/3) for s, and h = 2s/3.

Check my arithmetic.

Answer 2

Trying this without using Lagrange multipiers.
Constraint:
Volume, V = hs² = 72

Cost:
Cost, C = 40s²+4*30sh

so
h = 72/s^2
C = 40 s^2 + 120 s(72/s^2)
C = 40 s^2 + 8640/s
dC/ds = 0 at min = 80 s - 8640/s^2
so
80 s^3 = 8640 at min
s^3 = 108 = 27 * 4
so s = 3 * 4^(1/3) agrees with Math Mate
s = 4.76