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anonymous
 5 years ago
An open top rectangular box must have a volume of 4500 cu ft. If the bottom is a rectangle with length is twice the width, what dimensions would minimize total surface area of the box?
anonymous
 5 years ago
An open top rectangular box must have a volume of 4500 cu ft. If the bottom is a rectangle with length is twice the width, what dimensions would minimize total surface area of the box?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0V = l w h SA = 2(lw + lh + wh) l = 2w V = 2 w w h SA = 2(2w w + 2wh + wh) 4500 = 2 w w h h = 4500 / 2 w w SA = 2(2 w w + 2 w (4500 / 2 w w ) + w (4500 / 2 w w )) SA = 4*w w+13500*w w w SA' = ... Minimize it simply
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