Hello, 2taall!

We use the same procedure of Minimization as for Maximization.

The tricky part (of course) is setting up the function to optimize.

I'll walk you thorough the set-up.

A rectangular box is to have a square base and a volume of 20 ft³.

The material for the base costs 30˘/ft², for the sides 10˘/ft², and the top 20˘/ft².

Determine the dimensions of the box that can be constructed at minimum cost. Code:

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The volume of the box is: .$\displaystyle x^2y \:=\:20\quad\Rightarrow\quad y \:=\:\frac{20}{x^2}$ .**[1]**

The base has area $\displaystyle x^2$ ft². .Its cost is: .$\displaystyle 30x^2$ cents.

The top has area $\displaystyle x^2$ ft². .Its cost is: .$\displaystyle 20x^2$ cents.

The sides have area $\displaystyle 4xy$ ft². .Their cost is: .$\displaystyle 40xy$ cents.

Hence, the total cost is: .$\displaystyle C \;=\;30x^2 + 20x^2 + 40xy\quad\Rightarrow\quad C \:=\:50x^2 + 40xy $ .**[2]**

Substitute [1] into [2]: .$\displaystyle C \:=\:50x^2 + 40x\left(\frac{20}{x}\right)$

Hence, we must minimize: .$\displaystyle C \;=\;50x^2 + 800x^{-1}$