Cut 16" Pizza into 3 Equal Pieces Using Calculus

[MarkFL]

Here is the question:

How do you cut a 16 inch diameter pizza into 3 equal pieces using calculus?


It's a long one and someone told me 3pi/2 is the answer...I really need to know how to get that. Are there any similar problems on the internet I can follow to this one also?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Topher L,

One way to cut the pizza into 3 equal pieces is to make 3 radial cuts to divide the pizza into 3 equal sectors each subtending an angle of \(\displaystyle \frac{2\pi}{3}\). We simply divide the complete circle's angle of $2\pi$ by 3.

However, since you state that calculus is to be used, I suspect we are to make 2 parallel cuts along chords of the circle. So, if we orient the circle with its center at the origin of our $xy$-coordinate axes, we may cut along the lines $x=\pm c$. To find the value of $c$, we may state (where $r$ is the radius of the circle):

\(\displaystyle \int_0^c \sqrt{r^2-x^2}\,dx=\frac{\pi r^2}{12}\)

If we let:

\(\displaystyle x=r\sin(\theta)\,\therefore\,dx=r\cos(\theta)\)

We obtain:

\(\displaystyle r^2\int_0^{\sin^{-1}\left(\frac{c}{r} \right)} \cos^2(\theta)\,d\theta=\frac{\pi r^2}{12}\)

Using a double-angle identity for cosine, we may write:

\(\displaystyle \frac{r^2}{4}\int_0^{\sin^{-1}\left(\frac{c}{r} \right)} 1+\cos(2\theta)\,2\,d\theta=\frac{\pi r^2}{12}\)

Let $u=2\theta\,\therefore\,du=2\,d\theta$ and we have:

\(\displaystyle \frac{r^2}{4}\int_0^{2\sin^{-1}\left(\frac{c}{r} \right)} 1+\cos(u)\,du=\frac{\pi r^2}{12}\)

Using the anti-derivative and the FTOC, there results:

\(\displaystyle \frac{r^2}{4}\left[u+\sin(u) \right]_0^{2\sin^{-1}\left(\frac{c}{r} \right)}=\frac{\pi r^2}{12}\)

\(\displaystyle \frac{r^2}{4}\left(2\sin^{-1}\left(\frac{c}{r} \right)+\sin\left(2\sin^{-1}\left(\frac{c}{r} \right) \right) \right)=\frac{\pi r^2}{12}\)

Multiply through by \(\displaystyle \frac{12}{r^2}\):

\(\displaystyle 3\left(2\sin^{-1}\left(\frac{c}{r} \right)+\sin\left(2\sin^{-1}\left(\frac{c}{r} \right) \right) \right)=\pi\)

Using the double-angle identity for sine, we have:

\(\displaystyle 6\left(\sin^{-1}\left(\frac{c}{r} \right)+\sin\left(\sin^{-1}\left(\frac{c}{r} \right) \right) \cos\left(\sin^{-1}\left(\frac{c}{r} \right) \right) \right)=\pi\)

\(\displaystyle 6\left(\sin^{-1}\left(\frac{c}{r} \right)+\frac{c\sqrt{r^2-c^2}}{r^2} \right)=\pi\)

We may arrange this as:

\(\displaystyle f(c)=6\left(\sin^{-1}\left(\frac{c}{r} \right)+\frac{c\sqrt{r^2-c^2}}{r^2} \right)-\pi=0\)

Without loss of generality, we may let the radius of the circle be 1 unit:

\(\displaystyle f(c)=6\left(\sin^{-1}(c)+c\sqrt{1-c^2} \right)-\pi=0\)

Using a numeric root-finding technique, we find:

\(\displaystyle c\approx0.264932084602777\)

Since the radius of the pizza in the given problem is 8 inches, we then find that the cuts should be made along the lines:

\(\displaystyle c\approx\pm2.119456676822216\)

 

[Tennisgoalie]

At one point, you suddenly replaced c with sin^-1 (c/r). Can you explain why you did that?

 

[MarkFL]

At one point, you suddenly replaced c with sin^-1 (c/r). Can you explain why you did that?

Yes, I did so in accordance with the substitution I made. Let's go back to this point:

\(\displaystyle \int_0^c \sqrt{r^2-x^2}\,dx=\frac{\pi r^2}{12}\)

Now, we next used the substitution:

\(\displaystyle x=r\sin(\theta)\,\therefore\,dx=r\cos(\theta)\)

Now, originally the limits of integration are in terms of $x$, but we now want them to be in terms of the new variable $\theta$, and we find that:

\(\displaystyle x=r\sin(\theta)\)

May be solved for $\theta$ by dividing through by $r$ and arranging as:

\(\displaystyle \sin(\theta)=\frac{x}{r}\)

And this implies that we may write $\theta$ as a function of $x$ as follows:

\(\displaystyle \theta(x)=\sin^{-1}\left(\frac{x}{r} \right)\)

Hence, we find:

\(\displaystyle \theta(0)=\sin^{-1}\left(\frac{0}{r} \right)=0\)

\(\displaystyle \theta(c)=\sin^{-1}\left(\frac{c}{r} \right)\)

And these are our limits in terms of $\theta$, allowing us the write the definite integral as:

\(\displaystyle r^2\int_0^{\sin^{-1}\left(\frac{c}{r} \right)} \cos^2(\theta)\,d\theta\)

Does this make sense?

 

[Tennisgoalie]

This makes perfect sense, thank you.