Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 13/Solution 1

A = 2 \pi r^2 + 2\pi r h = 2 \pi r(r+h) </math> and the volume as $$V=\pi r^{2}h.$$ Our goal is to find the maximum and minimum volume of the cylinder, so we must find a formula for the volume that depends on only one variable. To do this, we solve for $h$ in the expression for the area: $$h=\frac{A}{2\pi r}-r=\frac{A-2\pi r^2}{2\pi r}.$$ Substituting this expression into the formula for the volume we obtain $$V=\pi r^2\frac{A-2\pi r^2}{2\pi r}=\frac{Ar}{2}-\pi r^3$$ which depends only on the radius $r$ and the area $A$. Since the area is fixed, we have expressed the volume as a function of only one variable, the radius.

Let us determine the domain of possible values the radius can attain. Since both the radius and volume must be positive, we must find the values of $r$ for which $r\ge 0$ and $V(r)\ge 0$. Since $V(r)=r(A/2-\pi r^2)$, the zeros of $V$ are $r=0$ and $r=\pm \sqrt{A/2\pi }$. Hence, $V(r)$ is positive on $(0,\sqrt{A/2\pi })$ and negative on $(\sqrt{A/2\pi },\mathrm{\infty })$, so the domain of possible radii is $[0,\sqrt{A/2\pi }]$.

Differentiating our expression $V=V(r)$ with respect to $r$ gives $$\frac{dV}{dr}=\frac{A}{2}-3\pi r^2,$$ and setting $\frac{dV}{dr}=0$ we obtain $r=\pm \sqrt{A/6\pi }$. Only the positive value is in the interval $[0,\sqrt{A/2\pi }]$, so we discard the negative value. Thus, the three values of interest are $\sqrt{A/6\pi }$ and the end points of the interval $0$ and $\sqrt{A/2\pi }$. Evaluating $V(r)$ at these values we obtain $$V(0)=V(\sqrt{A/2\pi })=0\text{and}V(\sqrt{A/6\pi })=\frac{A\sqrt{A/6\pi }}{2}-\pi (\sqrt{A/6\pi }{)}^3=\frac{A^{3/2}}{2\sqrt{6\pi }}-\frac{A^{3/2}}{6\sqrt{6\pi }}=\frac{A^{3/2}}{3\sqrt{6\pi }}.$$ Thus, we see that the maximum volume of the cylinder is $V=\frac{A^{3/2}}{3\sqrt{6\pi }}$, occurring at $r=\sqrt{A/6\pi }$ and its minimum is $V=0$ occurring at $r=0$ and $r=\sqrt{A/2\pi }$.