calculus problem!

Question

A rectangular storage container with a lid is to have a volume of 12 m3. The length of its base is twice the width. Material for the base costs $6 per m2. Material for the sides and lid costs $12 per m2. Find the dimensions of the container which will minimize cost and the minimum cost.

base width =  m

base length =  m

height =  m

minimum cost = $

Answer 1

Let x = width. Then 2x = length. Since x(2x)h = 12, h = 6 / x²

C = cost = 6(area of base) + 12(area of sides) + 12(area of top)

C = 6(2x²) + 12(2xh + 4xh + 2x²) = 36x² + 72xh = 36x² + 432x^-1

dC/dx = 72x - 432x^-2 = (72x³ - 432) / x² = 0 when 72x³ = 432. So, x = ³√6

x = width = ³√6

length = 2x = 2 (³√6)

height = 6 / (³√6)² = ³√6

Answer 2

12 = hLw= h(2w)w = 2hw^2

h= 12/2w^2 = 6/w^2

surface area = A = 2(Lw +wh +Lh)= twice bottom and top + twice the other two sides

Cost = (6Lw) +3Lw + 2(3wh) +2(Lh) $6 for the bottom, $3 for top and the other 4 sides

replace L with 2w, replace h with 6/w^2

Cost = C(w) = 3(2w) +6w(6/w^2) + 2(2w)(6/w^2)

= 6w + 36/w +24/w

= 6w +60/w

take the derivative of Cost with respect to Width,

and set equal to zero

C'(w) = 6 -60/w^2 =0

multiply by w^2

6w^2 = 60

w^2 = 60/6 = 10

w = sqr10 = width = about 3.16 meters

L = 2w = 2sqr10 = Length= about 6.32 meters

h = 6/w^2 = 6/10 = 0.6 meter

cost minimizing dimensions are about

3.16 by 6.32 by .06 meters

or exactly

sqr10 by 2sqr10 by 6/10

with Volume = 20x6/10 = 12 m^3

minimum Cost = 6w +60/w

= 6sqr10 + 60/sqr10

= about 6(3.16) + 60/3.16

= about 18.96 + 18.97

= $37.93

the arithmetic gets a little tedious,

with mistake(s) very possible

No guarantees the above is

error free, but one clue is that

bottom costs more so bottom should be smaller with height relatively larger to minimize costs

yet at first glace they surprisingly look about the same, so could go either way on suspected accuracy.

on 2nd glance the h=6/10 is much smaller than L or w. Odds are the answer is correct or in the right range.

Still this is the basic method. Redo it and see what you come up with.

calculus' derivatives are used for optimization (maximizig or minimizing) problems

where you take the derivative and set it = 0