Minimize surface area
Question
You have been asked to design a can with a volume of $824{\mathrm{cm}}^3$ 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:
$$\begin{array}{l}r=◻\mathrm{cm}\\ h=◻\mathrm{cm}\end{array}$$

Step 1 ：The volume of a right circular cylinder is given by the formula $V=\pi r^{2}h$, where $r$ is the radius and $h$ is the height. We are given that $V=824c{m}^3$. So, we have $\pi r^{2}h=824$.

Step 2 ：The surface area of a right circular cylinder with a closed top is given by the formula $A=2\pi r^2+2\pi rh$, where $A$ is the surface area. We want to minimize this.

Step 3 ：We can express $h$ in terms of $r$ and $V$ from the volume equation, which gives $h=\frac{V}{\pi r^2}=\frac{824}{\pi r^2}$.

Step 4 ：Substitute $h$ into the surface area equation, we get $A=2\pi r^2+2\pi r\cdot \frac{824}{\pi r^2}=2\pi r^2+\frac{1648}{r}$.

Step 5 ：To find the minimum surface area, we take the derivative of $A$ with respect to $r$ and set it equal to zero. So, $A^{\prime }=4\pi r-\frac{1648}{r^2}=0$.

Step 6 ：Solving for $r$, we get $r^3=\frac{1648}{4\pi }=131.946$. So, $r=\sqrt[3]{131.946}=5.08cm$ (rounded to two decimal places).

Step 7 ：Substitute $r=5.08cm$ into the equation for $h$, we get $h=\frac{824}{\pi (5.08{)}^2}=12.16cm$ (rounded to two decimal places).

Step 8 ：So, the radius $r$ and height $h$ that would minimize the amount of material needed to construct this can are approximately $\boxed{{\displaystyle 5.08cm}}$ and $\boxed{{\displaystyle 12.16cm}}$, respectively.