Cutting a can from a metal sheet to maximize volume

Yes, there are two cases. In your case 1 the side of the square equals the circumference of the lid; in case 2 the side of the square equals the height of the can.

In case 2 the volume of the can is a function of $r^2$ alone, so volume is maximized when $r$ is as large as possible. This is the reason why no calculus is required. The value of $r$ in this case is constrained by $2r + 2\pi r\le 10$, as you've shown.

The other answers have addressed case 1. It remains to check that the max volume under case 2 is less than the max volume under case 1.

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Calculus

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