Answer
a. $x=12cm$, $y=6cm$
b. $x=12cm$, $y=6cm$
Work Step by Step
a. Using the figure in the Exercise, we have $2x+2y=36$ or $y=18-x$. Assuming the radius of the cylinder is $r$, we have $2\pi r=x$ or $r=\frac{x}{2\pi}$. The volume is given as $V=\pi r^2y=\pi (\frac{x}{2\pi})^2(18-x)=\frac{1}{4\pi}(18x^2-x^3)$, thus $V'=\frac{1}{4\pi}(36x-3x^2)$. Let $V'=0$ to get $x=0,12$. As $x\gt0$, choose $x=12cm$ which gives $y=6cm$. Check $V''=\frac{1}{4\pi}(36-6x)|_{x=12}\lt0$; thus the region is concave down with a maximum volume.
b. In this case, $V=\pi x^2y=\pi x^2(18-x)=\pi(18x^2-x^3)$ (this is the same situation as in part-a), thus $V'=\pi(36x-3x^2)$. Let $V'=0$ to get $x=0,12$. As $x\gt0$, choose $x=12cm$ which gives $y=6cm$. Check $V''=\pi(36-6x)|_{x=12}\lt0$; thus the region is concave down with a maximum volume.
