Wednesday

November 26, 2014

November 26, 2014

Posted by **Travis** on Wednesday, March 6, 2013 at 2:15pm.

Base length=?m

Base width=?m

height=?

minimum cost=$?

- Calculus -
**Steve**, Wednesday, March 6, 2013 at 2:30pmif 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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