Question

what are the dimensions of the lightest open - top right circular cylindrical can that will hold a volume of 10,648 cm³?
the radius of the can is □ cm and its height is □ cm.
(type exact answers, using π as needed.)

Explanation:

Step1: Recall the volume and surface area formulas

The volume of a cylinder \(V=\pi r^{2}h\), and the surface area of an open - top cylinder \(S = \pi r^{2}+2\pi rh\). Given \(V = 10648\), so \(h=\frac{V}{\pi r^{2}}=\frac{10648}{\pi r^{2}}\). Substitute \(h\) into the surface area formula: \(S(r)=\pi r^{2}+2\pi r\times\frac{10648}{\pi r^{2}}=\pi r^{2}+\frac{21296}{r}\).

Step2: Find the derivative of \(S(r)\)

Differentiate \(S(r)\) with respect to \(r\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(S^\prime(r)=2\pi r-\frac{21296}{r^{2}}\).

Step3: Set the derivative equal to zero

Set \(S^\prime(r) = 0\), then \(2\pi r-\frac{21296}{r^{2}}=0\). Multiply through by \(r^{2}\) to get \(2\pi r^{3}-21296 = 0\). So \(r^{3}=\frac{21296}{2\pi}=\frac{10648}{\pi}\). Taking \(r=\sqrt[3]{\frac{10648}{\pi}}\). Since \(10648 = 22^{3}\), then \(r=\frac{22}{\sqrt[3]{\pi}}\).

Step4: Find the height \(h\)

Substitute \(r\) into \(h=\frac{V}{\pi r^{2}}\). \(h=\frac{10648}{\pi\times(\frac{22}{\sqrt[3]{\pi}})^{2}}=\frac{10648}{\pi\times\frac{484}{\pi^{2/3}}}=\frac{10648\pi^{2/3}}{484\pi}=\frac{22}{\sqrt[3]{\pi}}\) (because \(10648\div484 = 22\)).

Answer:

The radius of the can is \(\frac{22}{\sqrt[3]{\pi}}\text{ cm}\) and its height is \(\frac{22}{\sqrt[3]{\pi}}\text{ cm}\).