# leprofece's question at Yahoo! Answers: minimizing the lateral area of a cone with fixed volume

#### MarkFL

Staff member
Here is the question:

252) That shape must have a circular cone of volume "V" so the lateral area is going to be minimal.?

Answer is sqrt Cubic of ( 6V/pi) H = sqrt of 2 = V2
I have posted a link there to this thread so the OP can view my work.

#### MarkFL

Staff member
Hello leprofece,

Our objective function, the lateral surface area of a right circular cone of base radius $r$ and height $h$ is:

$$\displaystyle f(h,r)=\pi r\sqrt{r^2+h^2}$$

And we are subject to the constraint:

$$\displaystyle g(h,r)=\frac{\pi}{3}hr^2-V=0$$

If we assume that $0<V$, then we must also assume the measures $h$ and $r$ are also positive.

If we solve the constraint for $h$, we obtain:

$$\displaystyle h=\frac{3V}{\pi r^2}$$

Now, substituting for $h$ into the objective function, we get:

$$\displaystyle f(r)=\pi r\sqrt{r^2+\left(\frac{3V}{\pi r^2} \right)^2}=\frac{\sqrt{\pi^2r^6+9V^2}}{r}$$

Next, we want to differentiate with respect to $r$ and equate the result to zero to determine the critical value(s). We may do so using the quotient rule:

$$\displaystyle f'(r)=\frac{r\left(\dfrac{6\pi ^2r^5}{2\sqrt{\pi^2r^4+9V^2}} \right)-\sqrt{\pi^2r^6+9V^2}}{r^2}=\frac{2\pi^2r^6-9V^2}{r^2\sqrt{\pi^2r^6+9V^2}}=0$$

This implies:

$$\displaystyle 2\pi^2r^6-9V^2=0$$

Solving for $r$, and taking the positive root, we find:

$$\displaystyle r^6=\frac{9V^2}{2\pi^2}$$

$$\displaystyle r=\sqrt{\frac{3V}{\sqrt{2}\pi}}$$

Observing that the denominator of the first derivative of the objective function will always be positive for $0<r$, we may look only at the numerator and observe that to the left of the critical value the derivative is negative and to the right it is positive, and so by the first derivative test, we may conclude that the critical value we found is associated with a local minimum of the objective function.

Hence, the measures of the cone of volume $V$ having minimal lateral surface area are:

$$\displaystyle h\left(\sqrt{\frac{3V}{\sqrt{2}\pi}} \right)=\frac{3V}{\pi \left(\sqrt{\frac{3V}{\sqrt{2}\pi}} \right)^2}=\sqrt{\frac{6V}{\pi}}=\sqrt{2}r$$

$$\displaystyle r=\sqrt{\frac{3V}{\sqrt{2}\pi}}$$