Maximizing Volume: Finding Optimal Dimensions for a Box with an Open Top

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SUMMARY

The discussion focuses on maximizing the volume of a box with a square base and an open top, constrained by a material area of 60 square inches. The volume function is defined as V = x²y, where x is the side length of the base and y is the height. The area constraint leads to the equation x² + 4xy = 60, allowing for the substitution of y to express volume solely in terms of x. The next step involves differentiating the volume function and finding critical points to determine the optimal dimensions for maximum volume.

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Homework Statement



A box with a square base is to have an open top. The area of the material in the box is 60 sr in. What are the dimensions when the volume is maximized?

Homework Equations





The Attempt at a Solution


A=x2+4xy
V=x2y

x2+4xy=60
4xy=60-x2
y=(60-x2)/4x

Where do I go from here?
 
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Have you done this kind of maximization problem before? You want to get an function for volume in terms of a single variable (which you are about to do). Then you differentiate the function and set the derivative equal to zero.
 

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