Good Question 10 – The Cone Problem

Today’s good question is an optimization problem, but its real point is choosing how to do the computation. As such it relates to MPAC 3a and 3b: “Students can … select appropriate mathematical strategies [and] sequence algebraic/computational processes logically.” The algebra required to solve this questions can be quite daunting, unless you get clever. Here’s the question.

A sector of arc length x is removed from a circle of radius 10 cm. The remaining part of the circle is formed into a cone of radius r and height h,

Find the value of x so that the cone has the maximum possible volume.

The sector that was removed is also formed into a cone. Find the value of x that makes this cone have it maximum possible volume. (Hint: This is an easy problem.)

In the context of the problem, the expression for the volume of the cone in part a. has a domain of $0\le x\le 20\pi$ . Why? Ignore the physical situation and determine the domain of the expression for the volume from a. Graph the function. Discuss.

Solutions:

Part a: As usual, we start by assigning some variables.

Let r be the radius of the base of the cone and let h be its height. The circumference of the cone is $2\pi r=20\pi -x$ , so $r=10-\frac{x}{2\pi }$ and $h=\sqrt{{{10}^{2}}-{{r}^{2}}}$ . The volume of the cone is

$\displaystyle V=\frac{\pi }{3}{{r}^{2}}h=\frac{\pi }{3}{{r}^{2}}\sqrt{{{10}^{2}}-{{r}^{2}}}=\frac{\pi }{3}{{\left( 10-\frac{x}{2\pi } \right)}^{2}}\sqrt{{{10}^{2}}-{{\left( 10-\frac{x}{2\pi } \right)}^{2}}}$

To find the maximum, the next step is to differentiate the volume. The expression on the right above looks way complicated and its derivative will be even worse. Simplifying it is also a lot of trouble, and, in fact, does not make things easier.* Here’s where we can be clever and avoid a lot of algebra. Let’s just work from $\displaystyle V=\frac{\pi }{3}{{r}^{2}}\sqrt{{{10}^{2}}-{{r}^{2}}}$

To find the maximum differentiate the volume with respect to x using the chain rule.

$\displaystyle \frac{dV}{dx}=\frac{dV}{dr}\cdot \frac{dr}{dx}=\frac{\pi }{3}\left( {{r}^{2}}\frac{-2r}{2\sqrt{{{10}^{2}}-{{r}^{2}}}}+2r\sqrt{{{10}^{2}}-{{r}^{2}}} \right)\left( -\frac{1}{2\pi } \right)$

Setting this equal to zero and simplifying (multiply by $-6\sqrt{{{10}^{2}}-{{r}^{2}}}$ ) gives

$-{{r}^{3}}+2r\left( 100-{{r}^{2}} \right)=200r-3{{r}^{3}}=0$

$\displaystyle r=0,r=\sqrt{\frac{200}{3}}=\frac{10\sqrt{6}}{3}$

The minimum is obviously r = 0, so the maximum occurs when $\displaystyle r=10-\frac{x}{2\pi }=\frac{10\sqrt{6}}{3}$ . Then, solving for x gives

$\displaystyle x=2\pi \left( 10-\frac{10\sqrt{6}}{3} \right)\approx 11.52986$

Aside: We often see questions saying, if y = f(u) and u = g(x), find dy/dx. Here we have put that idea to practical use to save doing a longer computation.

Part b: The arc of the piece cut out is the circumference, x, of a cone with a radius of $\displaystyle {{r}_{1}}=\frac{x}{2\pi }$ and a height of $\displaystyle {{h}_{1}}=\sqrt{{{10}^{2}}-{{r}_{1}}^{2}}$ . Its volume is

$\displaystyle V=\frac{\pi }{3}{{r}_{1}}^{2}\sqrt{{{10}^{2}}-{{r}_{1}}^{2}}$

This is the same as the expression we used in part a. and can be handled the same way, except that here $\displaystyle \frac{d{{r}_{1}}}{dx}=+\frac{1}{2\pi }$ . The computation and result will be the same. The result will be the same. The maximum occurs at

$\displaystyle x=2\pi \left( 10-\frac{10\sqrt{6}}{3} \right)\approx 11.52986$

This should not be a surprise. The piece cut out and the piece that remains are otherwise indistinguishable, so the maximum volume should be the same for both.

Part c: From part a we have $\displaystyle V=\frac{\pi }{3}{{r}^{2}}\sqrt{{{10}^{2}}-{{r}^{2}}}=\frac{\pi }{3}{{\left( 10-\frac{x}{2\pi } \right)}^{2}}\sqrt{{{10}^{2}}-{{\left( 10-\frac{x}{2\pi } \right)}^{2}}}$ . To graph there is no need to simplify the expression in x:

$Tthe x scale marks are at multiples of $latex 5\pi $$

The x-scale marks are at multiples of $5\pi$

The domain is determined by the expression under the radical so

$-10\le r\le 10$

$-10\le 10-\frac{x}{2\pi }\le 10$

$0\le x\le 40\pi$

This is the “natural domain” of the function without regard to the physical situation given in the original problem. I cannot think of a reason for the difference.

________________

*Fully simplified in terms of x the volume is $\displaystyle V=\frac{1}{24{{\pi }^{2}}}{{\left( 20\pi -x \right)}^{2}}\sqrt{40\pi x-{{x}^{2}}}$ . This isn’t really easier to differentiate and solve.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Good Question 10 – The Cone Problem

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