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The objective is to minimize the surface area. Drawing a diagram and decomposing the surface into two circles and a rectangle gives: The area of each circle is \(\displaystyle A_c = \pi r^2\) and the area of the rectangle is \(\displaystyle A_r = 2\pi r h\). The function to be minimized is \(\displaystyle f = 2\pi r^2+2\pi r h\). The constraint on the volume can be used to eliminate \(\displaystyle h\) from the function \(\displaystyle f\). Solving \(\displaystyle 1.5 = \pi r^2 h\) for \(\displaystyle h\) gives \(\displaystyle h = \frac{1.5}{\pi r^{2}}\)Substituting \(\displaystyle h\) into \(\displaystyle f\) gives \(\displaystyle f = 2\pi r^2+2\pi r \left( \frac{1.5}{\pi r^2}\right) = 2\pi r^2 + \frac{3.0}{r}\). Taking the derivative gives \(\displaystyle f'(r) = 4 \pi r - \frac{3.0}{r^{2}}\). Clearing the fractions gives \(\displaystyle 3.0=4 \pi r^{3} \iff r^3=0.75 \iff r = \frac{\sqrt[3]{0.75}}{\sqrt[3]{\pi}}\)Substituing back into the equation for \(\displaystyle h\) gives \(\displaystyle h = \sqrt[3]{\frac{6.0}{\pi}}\)

Download \(\LaTeX\)

\begin{question}A circular cylinder needs to hold a volume of 1.5 square feet. Find the dimensions that minimize the amount of material needed, that is, minimize the surface area. 
    \soln{18cm}{The objective is to minimize the surface area.  Drawing a diagram and decomposing the surface into two circles and a rectangle gives:
\begin{center}
\begin{tikzpicture}
	 \draw(0, 0) -- (0, 3);
	 \draw(2, 0) -- (2, 3);
	\draw[dashed, yscale=.25] (0, 0) arc (180:0:1);
	\draw[yscale=.25] (0, 0) arc (180:360:1);
	\draw[yscale=.25] (2, 12) arc (0:360:1);
	 \draw[latex-latex] (-0.2, 0) -- (-0.2, 3) node[midway, left]{$h$};
	 \draw[latex-latex] (0, 0) -- (1, 0) node[midway, yshift=-.35cm]{$r$};
		\draw (4,2.5) circle (28pt);
		\draw (4,0.5) circle (28pt);
		\draw(7, 0) rectangle (13, 3);
	 \draw[latex-latex] (7, -0.2) -- (13, -0.2) node[midway, yshift=-.35cm]{$2\pi r$};
	 \draw[latex-latex] (6.8, 0) -- (6.8, 3) node[midway, left]{$h$};
	 \draw[latex-latex] (4, 2.5) -- (5, 2.5) node[midway, yshift=-.35cm]{$r$};
	 \draw[latex-latex] (4, 0.5) -- (5, 0.5) node[midway, yshift=-.35cm]{$r$};
\end{tikzpicture}
\end{center}

The area of each circle is $A_c = \pi r^2$ and the area of the rectangle is $A_r = 2\pi r h$. The function to be minimized is $f = 2\pi r^2+2\pi r h$.  The constraint on the volume can be used to eliminate $h$ from the function $f$. Solving $1.5 = \pi r^2 h$ for $h$ gives $h = \frac{1.5}{\pi r^{2}}$Substituting $h$ into $f$ gives $f = 2\pi r^2+2\pi r \left( \frac{1.5}{\pi r^2}\right) = 2\pi r^2 + \frac{3.0}{r}$. Taking the derivative gives $f'(r) = 4 \pi r - \frac{3.0}{r^{2}}$. Clearing the fractions gives $3.0=4 \pi r^{3} \iff r^3=0.75 \iff r = \frac{\sqrt[3]{0.75}}{\sqrt[3]{\pi}}$Substituing back into the equation for $h$ gives $h = \sqrt[3]{\frac{6.0}{\pi}}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}

\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}

\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue

\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>A circular cylinder needs to hold a volume of 1.5 square feet. Find the dimensions that minimize the amount of material needed, that is, minimize the surface area. </p> </p>

HTML for Canvas

<p> <p>The objective is to minimize the surface area.  Drawing a diagram and decomposing the surface into two circles and a rectangle gives:<?xml version="1.0" encoding="UTF-8"?>
<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="612pt" height="792pt" viewBox="0 0 612 792" version="1.1">
<g id="surface1">
</g>
</svg>
The area of each circle is  <img class="equation_image" title=" \displaystyle A_c = \pi r^2 " src="/equation_images/%20%5Cdisplaystyle%20A_c%20%3D%20%5Cpi%20r%5E2%20" alt="LaTeX:  \displaystyle A_c = \pi r^2 " data-equation-content=" \displaystyle A_c = \pi r^2 " />  and the area of the rectangle is  <img class="equation_image" title=" \displaystyle A_r = 2\pi r h " src="/equation_images/%20%5Cdisplaystyle%20A_r%20%3D%202%5Cpi%20r%20h%20" alt="LaTeX:  \displaystyle A_r = 2\pi r h " data-equation-content=" \displaystyle A_r = 2\pi r h " /> . The function to be minimized is  <img class="equation_image" title=" \displaystyle f = 2\pi r^2+2\pi r h " src="/equation_images/%20%5Cdisplaystyle%20f%20%3D%202%5Cpi%20r%5E2%2B2%5Cpi%20r%20h%20" alt="LaTeX:  \displaystyle f = 2\pi r^2+2\pi r h " data-equation-content=" \displaystyle f = 2\pi r^2+2\pi r h " /> .  The constraint on the volume can be used to eliminate  <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX:  \displaystyle h " data-equation-content=" \displaystyle h " />  from the function  <img class="equation_image" title=" \displaystyle f " src="/equation_images/%20%5Cdisplaystyle%20f%20" alt="LaTeX:  \displaystyle f " data-equation-content=" \displaystyle f " /> . Solving  <img class="equation_image" title=" \displaystyle 1.5 = \pi r^2 h " src="/equation_images/%20%5Cdisplaystyle%201.5%20%3D%20%5Cpi%20r%5E2%20h%20" alt="LaTeX:  \displaystyle 1.5 = \pi r^2 h " data-equation-content=" \displaystyle 1.5 = \pi r^2 h " />  for  <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX:  \displaystyle h " data-equation-content=" \displaystyle h " />  gives  <img class="equation_image" title=" \displaystyle h = \frac{1.5}{\pi r^{2}} " src="/equation_images/%20%5Cdisplaystyle%20h%20%3D%20%5Cfrac%7B1.5%7D%7B%5Cpi%20r%5E%7B2%7D%7D%20" alt="LaTeX:  \displaystyle h = \frac{1.5}{\pi r^{2}} " data-equation-content=" \displaystyle h = \frac{1.5}{\pi r^{2}} " /> Substituting  <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX:  \displaystyle h " data-equation-content=" \displaystyle h " />  into  <img class="equation_image" title=" \displaystyle f " src="/equation_images/%20%5Cdisplaystyle%20f%20" alt="LaTeX:  \displaystyle f " data-equation-content=" \displaystyle f " />  gives  <img class="equation_image" title=" \displaystyle f = 2\pi r^2+2\pi r \left( \frac{1.5}{\pi r^2}\right) = 2\pi r^2 + \frac{3.0}{r} " src="/equation_images/%20%5Cdisplaystyle%20f%20%3D%202%5Cpi%20r%5E2%2B2%5Cpi%20r%20%5Cleft%28%20%5Cfrac%7B1.5%7D%7B%5Cpi%20r%5E2%7D%5Cright%29%20%3D%202%5Cpi%20r%5E2%20%2B%20%5Cfrac%7B3.0%7D%7Br%7D%20" alt="LaTeX:  \displaystyle f = 2\pi r^2+2\pi r \left( \frac{1.5}{\pi r^2}\right) = 2\pi r^2 + \frac{3.0}{r} " data-equation-content=" \displaystyle f = 2\pi r^2+2\pi r \left( \frac{1.5}{\pi r^2}\right) = 2\pi r^2 + \frac{3.0}{r} " /> . Taking the derivative gives  <img class="equation_image" title=" \displaystyle f'(r) = 4 \pi r - \frac{3.0}{r^{2}} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28r%29%20%3D%204%20%5Cpi%20r%20-%20%5Cfrac%7B3.0%7D%7Br%5E%7B2%7D%7D%20" alt="LaTeX:  \displaystyle f'(r) = 4 \pi r - \frac{3.0}{r^{2}} " data-equation-content=" \displaystyle f'(r) = 4 \pi r - \frac{3.0}{r^{2}} " /> . Clearing the fractions gives  <img class="equation_image" title=" \displaystyle 3.0=4 \pi r^{3} \iff r^3=0.75 \iff r = \frac{\sqrt[3]{0.75}}{\sqrt[3]{\pi}} " src="/equation_images/%20%5Cdisplaystyle%203.0%3D4%20%5Cpi%20r%5E%7B3%7D%20%5Ciff%20r%5E3%3D0.75%20%5Ciff%20r%20%3D%20%5Cfrac%7B%5Csqrt%5B3%5D%7B0.75%7D%7D%7B%5Csqrt%5B3%5D%7B%5Cpi%7D%7D%20" alt="LaTeX:  \displaystyle 3.0=4 \pi r^{3} \iff r^3=0.75 \iff r = \frac{\sqrt[3]{0.75}}{\sqrt[3]{\pi}} " data-equation-content=" \displaystyle 3.0=4 \pi r^{3} \iff r^3=0.75 \iff r = \frac{\sqrt[3]{0.75}}{\sqrt[3]{\pi}} " /> Substituing back into the equation for  <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX:  \displaystyle h " data-equation-content=" \displaystyle h " />  gives  <img class="equation_image" title=" \displaystyle h = \sqrt[3]{\frac{6.0}{\pi}} " src="/equation_images/%20%5Cdisplaystyle%20h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B6.0%7D%7B%5Cpi%7D%7D%20" alt="LaTeX:  \displaystyle h = \sqrt[3]{\frac{6.0}{\pi}} " data-equation-content=" \displaystyle h = \sqrt[3]{\frac{6.0}{\pi}} " /> </p> </p>