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: \[ \begin{array}{l} r=\square \mathrm{cm} \\ h=\square \mathrm{cm} \end{array} \]

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 \(V = 824 \, cm^3\). 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, \(A' = 4\pi r - \frac{1648}{r^2} = 0\).

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

Step 7 ：Substitute \(r = 5.08 \, cm\) into the equation for \(h\), we get \(h = \frac{824}{\pi (5.08)^2} = 12.16 \, cm\) (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 \(\boxed{5.08 \, cm}\) and \(\boxed{12.16 \, cm}\), respectively.