MHB Cut 16" Pizza into 3 Equal Pieces Using Calculus

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To cut a 16-inch diameter pizza into three equal pieces using calculus, one approach involves making three radial cuts, each subtending an angle of 2π/3. Alternatively, two parallel cuts along chords can be used, requiring integration to find the correct position of the cuts. The integration involves setting up an equation to ensure each piece has equal area, leading to a numeric solution for the cut positions. The calculated value for the cuts is approximately ±2.12 inches from the center. This method effectively demonstrates the application of calculus in solving geometric problems.
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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.
 
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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 $$\frac{2\pi}{3}$$. 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):

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

If we let:

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

We obtain:

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

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

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

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

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

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

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

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

Multiply through by $$\frac{12}{r^2}$$:

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

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

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

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

We may arrange this as:

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

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

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

Using a numeric root-finding technique, we find:

$$c\approx0.264932084602777$$

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:

$$c\approx\pm2.119456676822216$$
 
At one point, you suddenly replaced c with sin^-1 (c/r). Can you explain why you did that?
 
Tennisgoalie said:
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:

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

Now, we next used the substitution:

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

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:

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

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

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

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

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

Hence, we find:

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

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

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

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

Does this make sense?
 
This makes perfect sense, thank you.
 
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