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effort
Senior Member

USA
38 Posts

Posted - 09/09/2013 :  11:39:29  Show Profile  Reply with Quote
A builder intends to construct a storage shed having a volume of V ft, a flat roof, and a rectangular base whose width is three-fourths the length. the cost per square foot of the material is $4 for the floor, $6 for the sides, and $3 for the roof. What dimensions will
minimize the builder's cost?


[blue]Here is what I have so far but now I am stuck. Will someone check my work and show me how to finish solving this?[blue]

area of floor= 4wl, shed has 4 sides, 2 with area length times height and 2 with area width times height. so the cost of the material is
6lh and 6(2wh). the cost of the roof is 3wl.

Cost = 4(L)(3L/4)+ 3(L)(3L/4) + 6*2(L + 3L/4)h
= 3L + 9L/4 +21Lh

= 21L/4 + 21Lh
Now I know I am suppose to sub in a valus for a variable and then differentiate but I don't have a value to use to sub in. So I can't find the dimensions. Will someone show me how to finish this problem?
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royhaas
Moderator

USA
3059 Posts

Posted - 09/09/2013 :  13:38:16  Show Profile  Reply with Quote
Since the volume is constant, the height is the volume divided by the area of the rectangular base.
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