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

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MATH100 B December 2024

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Question 13
Consider a cylinder. If its total surface area, including both its top and bottom, is A, what is the largest possible volume it can enclose? What is the smallest possible volume? Your answers may depend on the constant A.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Write formulas for the surface area and volume of a cylinder in terms of its radius $r$ and height $h$. Use the surface area constraint to eliminate one variable, so that the volume is expressed as a function of a single variable.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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 $r=\sqrt{A/2\pi }$.

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