maximize volume problem, please help!?

Question

A closed rectangular box with a square base is to be built for a cost of $60. The material for the base costs $3 per square foot, the material for the top costs $2 per square foot, and the material for the sides costs $1 per square foot. What is the maximum volume of such a box?

Answer 1

4s^2 + 8sh = 60
V = hs^2
h = (15 - s^2)/2s
V = (1/2)(15s - s^3)
dV/ds = (1/2)(15 - 3s^2) = 0 for max volume
3s^2 = 15
s = √5
h = (15 - 5)/(2√5) = √5
V = 5√5 = 11.18 ft^3

Answer 2

w = length of each side for square base (and top)
h = height of box

Volume of box = V = hw^2

Area of Top = Area of Base = w^2
Area of Sides = 4wh

Cost = Cost of Top + Cost of Base + Cost of Sides = 60

(2)w^2 + (3)w^2 + (1)4wh = 60

5w^2 + 4wh - 60 = 0

This quadratic equation has two roots, a positive one and a negative one.....obviously, we can't have a negative value for dimension, so w is equal to the positive root, which is:

w = [-b + (b^2 - 4ac)^0.5]/2a (a = 5, b = 4h, c = -60)

w = [-4h + (16h^2 + 1200)^0.5]/10

= [-4h + 4(h^2 + 75)^0.5]/10

= [-2h + 2(h^2 + 75)^0.5]/5

Normally, the next step would be to write the above Volume expression (V = hw^2) in terms of h, take the first derivative (V'), and set it equal to 0. The value of h that makes this condition true is the dimension that will yield the maximum volume, but rather than dealing with the first derivative (which looks a little ugly and unwieldy in this case), we can just do a little trial and error for the optimum values of h (and w). The simplest and most efficient way to do this is with a program like MS Excel.

Personally, I tested values for h from 1 to 10, got a corresponding figure for w in each case, then calculated the resulting volume from those 2 numbers. If you repeat this process yourself, you will see that the volume hits a maximum at the following dimensions:

h = 5 ft; w = 2 ft (V = 5*2^2 = 20 ft^3)