Answer to Question #108391 in Calculus for Prabhjot

Question

Find the dimensions of a right circular cylinder that is open on the top, is closed on the bottom, has volume V , and uses the least amount of material.

Answer 1

Solution:

Consider the right circular cylinder of radius $r$ and height $h$.

Given, the volume of the cylinder is $V$ , so

$$\pi r^2h=V$$ $$h=\frac{V}{\pi r^2}$$

Total surface area of cylinder(excluding top) is given by,

$$A=2\pi rh+\pi r^2$$

Substitute $\frac{V}{\pi r^2}$ for $h$ into equation $A=2\pi rh+\pi r^2$ to express the area in terms of single variable $r$ as,

$$A(r)=2\pi r(\frac{V}{\pi r^2})+\pi r^2$$ $$A(r)=\frac{2V}{ r}+\pi r^2$$

Differentiate $A(r)$ with respect to $r$ as,

$$A'(r)=\frac{d}{dr}(\frac{2V}{ r}+\pi r^2)=-\frac{2V}{r^2}+2\pi r$$

Equate the derivative $A'(r)$ to zero to obtain,

$$-\frac{2V}{r^2}+2\pi r=0$$ $$r=\sqrt[3]{\frac{V}{\pi}}$$

Find second derivative to check for maxima/minima as,

"A"(r)=\\frac{d}{dr}(-\\frac{2V}{r^2}+2\\pi r)=\\frac{4V}{r^3}+2\\pi"

Here, "A"(r)=\\frac{4V}{r^3}+2\\pi" is positive for any positive value of $r$. So, minimum occurs at $r=\sqrt[3]{\frac{V}{\pi}}$ .

Now, plug $r=\sqrt[3]{\frac{V}{\pi}}$ into relation $h=\frac{V}{\pi r^2}$ and simplify for $h$ as,

$$h=\frac{V}{\pi (\frac{V^\frac{2}{3}}{\pi^\frac{2}{3}})}=\frac{V^\frac{1}{3}}{\pi ^\frac{1}{3}}=\sqrt[3]{\frac{V}{\pi}}$$

Therefore, the dimension of right circular cylinder is: Radius$(r)=\sqrt[3]{\frac{V}{\pi}}$ and Height$(h)=\sqrt[3]{\frac{V}{\pi}}$ .