Hello, Tactified!

I'll give you a complete walk-through . . . with baby-steps.

A closed rectangular box with a square base is to have a volume of 2000 cm³.

It costs twice as much per cm² for the top and bottom as it does for the sides.

Find the dimensions of the container of least cost. Code:

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The volume of the box is 2000 cm³: .$\displaystyle V \:=\:x^2y \:=\:2000 \quad\Rightarrow\quad y \:=\:\frac{2000}{x^2}$ .[1]

The four sides of the box have an area of $\displaystyle xy$ cm² each: .$\displaystyle \text{Area} \:=\:4xy$

Let $\displaystyle p$ = price of material per cm² for the sides.

. . $\displaystyle C_{\text{sides}} \:=\:4pxy$ dollars.

The ends (top and bottom) of the box have an area of $\displaystyle x^2$ cm² each: .$\displaystyle \text{Area} \:=\:2x^2$

The ends cost $\displaystyle 2p$ per cm².

. . $\displaystyle C_{\text{ends}} \:=\:4px^2$ dollars.

Hence, the total cost is: .$\displaystyle C \;=\;4px^2 + 4pxy$ .[2]

Substitute [1] into [2]: .$\displaystyle C \;=\;4px^2 + 4px\left(\frac{2000}{x^2}\right)$

Therefore, we must minimize: .$\displaystyle C\;=\;4px^2 + 8000px^{-1}$

Got it?

Awww, too slow again . . . *sigh*

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