# Calculus I

A square-bottomed box with no top has a fixed volume, V. What dimensions minimize the surface area?

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1. let each side of the base be x
let the height be y
volume is V, a constant
x^y = V
y = V/x^2

SA = x^2 + 4xy
= x^2 + 4xV/c^2 = x^2 + 4V/x
d(SA)/dx = 2x - 4V/x^2 = 0 for min of SA

4V/x^2 = 2x
x^3 = 2V
x = (2V)^(1/3)
take it from there

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posted by Reiny

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