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 #2
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Jan 10, 2019
 #1
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the wording is kind of odd. Is the bottom open? Sort of a weird way to orient the problem if soI'm going to assume both the top and bottom are closedwe've got the two different types of materials so let's compute the two areas separatelyassume the box has square side length s and height hareaside=4shareatb=2s2vol=s2h

 

we can convert areas into cost using the materials pricesC=4sh(1.5)+2s2(3)=6sh+6s2

 

I"m going to assume you haven't been taught Lagrange multipliers yetso we have to use the cost function to write the volume in a single variable15=6sh+6s2h=156s26svol=s2h=s2156s26s=15s6s36

 

Now to find the extrema of the volume we set it's derivative equal to 0dvolds=1518s26=56s22=05=6s2s=56 ft (we can ignore the negative solution)

 

vol=16(15566(56)3/2)=5356 ft3

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Jan 10, 2019
Jan 9, 2019
 #3
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Jan 9, 2019

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