Question

A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible surface area of such a cylinder.

Short answer

The maximum surface area of the cylinder is \(\pi {r^2}\left( {1 + \sqrt 5 } \right)\).

Step-by-step solution

Step 1-Find the function for total surface area

The figure below represents the sketch of the cylinder inside the sphere.

The total surface area of the cylinder is:

\(\begin{aligned}{c}A = 2\pi {y^2} + 2\pi y\left( {2x} \right)\\ = 2\pi \left( {{y^2} + 2yx} \right)\end{aligned}\)

The equation of the projected circle is:

\({x^2} + {y^2} = {r^2}\)

So, the equation of the total surface area of the cylinder is:

\(A = 2\pi \left[ {\left( {{r^2} - {x^2}} \right) + 2x\sqrt {{r^2} - {x^2}} } \right]\)

Step 2-Differentiate the function of total surface area

Differentiate the function \(A = 2\pi \left[ {\left( {{r^2} - {x^2}} \right) + 2x\sqrt {{r^2} - {x^2}} } \right]\).

\[\begin{aligned}{c}A' = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {2\pi \left[ {\left( {{r^2} - {x^2}} \right) + 2x\sqrt {{r^2} - {x^2}} } \right]} \right)\\ = 2\pi \left[ { - 2x + 2\left\{ {x\frac{1}{{2\sqrt {{r^2} - {x^2}} }}\left( { - 2x} \right) + \sqrt {{r^2} - {x^2}} \left( 1 \right)} \right\}} \right]\\ = 4\pi \left[ { - x - \frac{{{x^2}}}{{\sqrt {{r^2} - {x^2}} }} + \sqrt {{r^2} - {x^2}} } \right]\\ = 4\pi \cdot \frac{{ - x\sqrt {{r^2} - {x^2}} - 2{x^2} + {r^2}}}{{\sqrt {{r^2} - {x^2}} }}\end{aligned}\]

Step 3-Find the values of x at which the total surface area is maximum

Find the roots of \(A'\left( x \right) = 0\).

\[\begin{aligned}{c}4\pi \cdot \frac{{ - x\sqrt {{r^2} - {x^2}} - 2{x^2} + {r^2}}}{{\sqrt {{r^2} - {x^2}} }} = 0\\ - x\sqrt {{r^2} - {x^2}} - 2{x^2} + {r^2} = 0\\{x^2}\left( {{r^2} - {x^2}} \right) = {\left( {{r^2} - 2{x^2}} \right)^2}\\{r^2}{x^2} - {x^4} = {r^4} - 4{r^2}{x^2} + 4{x^4}\\5{x^4} - 5{r^2}{x^2} + {r^4} = 0\\{x^2} = \frac{{5 \pm \sqrt 5 }}{{10}}{r^2}\end{aligned}\]

The positive value of \({x^2}\) must is neglected, as it does not lie in the interval \(0 \le x \le r\).

Therefore, \(x = \sqrt {\frac{{5 - \sqrt 5 }}{{10}}} r\).

So, according to the first derivative test, the value of the surface area will be maximum when \(x = \sqrt {\frac{{5 - \sqrt 5 }}{{10}}} r\).

Step 4-Find the maximum total surface area

Substitute \(\sqrt {\frac{{5 - \sqrt 5 }}{{10}}} r\) for x in the equation \(A = 2\pi \left[ {\left( {{r^2} - {x^2}} \right) + 2x\sqrt {{r^2} - {x^2}} } \right]\).

\(\begin{aligned}{c}A = 2\pi \left[ {\left( {{r^2} - {x^2}} \right) + 2x\sqrt {{r^2} - {x^2}} } \right]\\ = 2\pi \left[ {\left( {{r^2} - \left( {\frac{{5 - \sqrt 5 }}{{10}}} \right){r^2}} \right) + 2\sqrt {\frac{{5 - \sqrt 5 }}{{10}}} r\sqrt {{r^2} - \left( {\frac{{5 - \sqrt 5 }}{{10}}} \right){r^2}} } \right]\\ = 2\pi {r^2}\left[ {\frac{{5 + \sqrt 5 }}{{10}} + 2\sqrt {\frac{{5 - \sqrt 5 }}{{10}}} \sqrt {\frac{{5 + \sqrt 5 }}{{10}}} } \right]\\ = 2\pi {r^2}\left[ {\frac{{5 + \sqrt 5 }}{{10}} + 2\frac{{\sqrt {25 - 5} }}{{10}}} \right]\\ = \pi {r^2}\left[ {\frac{{5 + \sqrt 5 }}{5} + \frac{{4\sqrt 5 }}{5}} \right]\\ = \pi {r^2}\left[ {\frac{{5 + 5\sqrt 5 }}{5}} \right]\\ = \pi {r^2}\left( {1 + \sqrt 5 } \right)\end{aligned}\)

Thus, the maximum surface area of the cylinder is \(\pi {r^2}\left( {1 + \sqrt 5 } \right)\).