Question

question 9 · 1 point you have been asked to design a can with a volume of 672 cm³ that is shaped like a right - circular cylinder. the can will have a closed top. what radius r and height h, in centimeters, would minimize the amount of material needed to construct this can? enter an exact answer. provide your answer below: r = cm h = cm

Explanation:

Step1: Recall volume and surface - area formulas

The volume formula for a right - circular cylinder is $V=\pi r^{2}h$, and we know $V = 672$, so $\pi r^{2}h=672$, which gives $h=\frac{672}{\pi r^{2}}$. The surface - area formula for a closed - top cylinder is $A = 2\pi r^{2}+2\pi rh$.

Step2: Substitute $h$ into the surface - area formula

Substitute $h=\frac{672}{\pi r^{2}}$ into $A$:
\[

$$\begin{align*} A(r)&=2\pi r^{2}+2\pi r\cdot\frac{672}{\pi r^{2}}\\ &=2\pi r^{2}+\frac{1344}{r} \end{align*}$$

\]

Step3: Find the derivative of $A(r)$

Using the power rule, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$.
\[

$$\begin{align*} A^\prime(r)&=4\pi r-\frac{1344}{r^{2}} \end{align*}$$

\]

Step4: Set the derivative equal to zero and solve for $r$

\[

$$\begin{align*} 4\pi r-\frac{1344}{r^{2}}&=0\\ 4\pi r&=\frac{1344}{r^{2}}\\ 4\pi r^{3}&=1344\\ r^{3}&=\frac{1344}{4\pi}=\frac{336}{\pi}\\ r&=\sqrt[3]{\frac{336}{\pi}} \end{align*}$$

\]

Step5: Find the value of $h$

Substitute $r = \sqrt[3]{\frac{336}{\pi}}$ into $h=\frac{672}{\pi r^{2}}$:
\[

$$\begin{align*} h&=\frac{672}{\pi(\sqrt[3]{\frac{336}{\pi}})^{2}}\\ &=\frac{672}{\pi\cdot(\frac{336}{\pi})^{\frac{2}{3}}}\\ &=2\sqrt[3]{\frac{336}{\pi}} \end{align*}$$

\]

Answer:

$r=\sqrt[3]{\frac{336}{\pi}}\text{ cm}$, $h = 2\sqrt[3]{\frac{336}{\pi}}\text{ cm}$