Calculus
posted by Travis .
A rectangular storage container with an open top is to have a volume of 10m^3. The length of its base is twice the width. Material for the base costs $3 per m^2. Material for the sides costs $10.8 per m^2. Find the dimensions of the container which will minimize cost and the minimum cost.
Base length=?m
Base width=?m
height=?
minimum cost=$?

if the base width is w,
base is w*2w
height is 10/2w^2 = 5/w^2
area is thus
a = w*2w + 2(w*h) + 2(2w*h)
= 2w^2 + 6wh
cost is thus
c = 3*2w^2 + 10.8*6wh
= 6w^2 + 64.8w*5/w^2
= 6w^2 + 324/w
dc/dw = 12w  324/w^2
= 12(w^3  27)/w^2
since w≠0,
dc/dw is minimum when w^3=27, or w=3
So, the box is 3 x 6 x 5/9
and the minimum cost is $162
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