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March 4th, 2017, 07:20 AM   #1
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Maximization Problem (rectangle inside of circular segment)

"A circle of radius a is divided into two segments by a line L at a distance b from the center. The rectangle of greatest possible area is inscribed in the smaller of those segments. How far from the center is the side of this rectangle that is opposite to the line L?"

Hi all. I'm trying to solve this problem, but am having a hard time finding the equation for the rectangle.

Here is a picture I made to visualize the problem:



So I realize that all I need to do to solve the problem is find the max. possible area of the rectangle inscribed in the smaller segment. The dimensions of the rectangle should be $\displaystyle (2x)*(y-b)$, right? Here's another picture:



I'm basically setting this up exactly like I set up (and correctly solved) the maximization of a rectangle inside of a circle/semicircle. But the -b term is causing me problems when I differentiate.

Here's my method/issue...

First, rewrite the area of the rectangle in terms of one variable. Since we know $\displaystyle x^2 + y^2 = r^2$ for any point on the circle (or circle segment), the area is: $\displaystyle 2x*[\sqrt(r^2-x^2)-b]$.

When I differentiate (and set to zero to maximize), I get:

$\displaystyle (-2x^2)/\sqrt(r^2-x^2)+2\sqrt(r^2-x^2)-2b=0$

I can't simplify this to solve for x!

Am I setting up this problem wrong? Or making some mistake in differentiating?

Last edited by skipjack; March 4th, 2017 at 09:34 AM.
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March 4th, 2017, 12:35 PM   #2
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You've done everything correctly as far as I can see.

You get 4 roots.

When you take into account $r>0,~0<b<r,~0<x<r$ you end up with

$x =\dfrac{\sqrt{-\sqrt{b^2 \left(b^2+8 r^2\right)}-b^2+4 r^2}}{2 \sqrt{2}}$

as the root that produces the maximum positive area.
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March 4th, 2017, 01:02 PM   #3
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derivative set equal to 0 ...

$\dfrac{-x^2}{\sqrt{r^2-x^2}} + \sqrt{r^2-x^2} - b = 0$

Quote:
How far from the center is the side of this rectangle that is opposite to the line L?
... that would be the value of $y$ on the circle that maximizes the rectangle's area, correct?

$\dfrac{y^2-r^2}{y} + y - b = 0$

$y - \dfrac{r^2}{y} + y - b = 0$

$2y - \dfrac{r^2}{y} - b = 0$

$2y^2 - by - r^2 = 0$

$y = \dfrac{b \pm \sqrt{(-b)^2 - 4(2)(-r^2)}}{2(2)}$

$y = \dfrac{b \pm \sqrt{b^2+8r^2}}{4}$

since $y > 0$ and $\sqrt{b^2+8r^2} > b$ ...

$y = \dfrac{b + \sqrt{b^2+8r^2}}{4}$


I tested this with $r=7$ and $b=5$

$A = 2x(\sqrt{7^2-x^2} - 5)$ with its max value is shown in the attached graph (note y in the graph represents the max area, not the y-value for the rectangle)

$y = \dfrac{5+\sqrt{5^2+8(7^2)}}{4} \approx 6.355144464$

$x = \sqrt{7^2-y^2} \approx 2.934644585$

$A = 2x(y-5) \approx 7.953734727$
Attached Images
File Type: png max_area.png (1.3 KB, 20 views)
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March 4th, 2017, 03:57 PM   #4
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I would let the objective function be:

$\displaystyle f(x,y)=xy$

Subject to the constraint:

$\displaystyle g(x,y)=x^2+(y+b)^2-r^2=0$

$\displaystyle 0<b<r$

Using Lagrange Multipliers, we obtain the system:

$\displaystyle y=\lambda(2x)$

$\displaystyle x=\lambda(2(y+b))$

This implies:

$\displaystyle \frac{y}{2x}=\frac{x}{2(y+b)}$

$\displaystyle x^2=y(y+b)$

Substituting into the constraint, we get:

$\displaystyle y(y+b)+(y+b)^2-r^2=0$

Taking the positive root, we obtain:

$\displaystyle y=\frac{\sqrt{b^2+8r^2}-3b}{4}$
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March 5th, 2017, 01:40 AM   #5
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Quote:
Originally Posted by romsek
you end up with ... as the root that produces the maximum positive area.
Thanks for checking! I was on the road to getting there when I thought that the x root was looking way too ugly... guess that was silly of me to think (I'm just so used to nice textbook answers).

Quote:
Originally Posted by skeeter
... that would be the value of y on the circle that maximizes the rectangle's area, correct?
Yes that should be right. The maximum y value of $\displaystyle 2x(y-b)$ should be the distance from the center to the farthest side of the rectangle (from the center/line L).

I like your solution a lot, making things a lot easier just cutting straight to the chase (to y). Thanks for showing me

Quote:
Originally Posted by MarkFL
I would let the objective function be:...
Thanks for a different sort of solution! I didn't know anything about Lagrange multipliers (or contours/gradient vectors) but I watched a few videos about how to use them until I could figure out your answer. I was freaking out about your numerator being 4b off from skeeter's until I realized that you had defined y as the height of the rectangle and that I needed to add in a factor of b (or 4b/b) to get the final answer (which matches skeeter's).

Thanks for a different sort of solution/sending me down a rabbit hole

thanks guys
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March 5th, 2017, 06:12 AM   #6
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Quote:
Originally Posted by rmoney View Post
Thanks for a different sort of solution! I didn't know anything about Lagrange multipliers (or contours/gradient vectors) but I watched a few videos about how to use them until I could figure out your answer. I was freaking out about your numerator being 4b off from skeeter's until I realized that you had defined y as the height of the rectangle and that I needed to add in a factor of b (or 4b/b) to get the final answer (which matches skeeter's).

Thanks for a different sort of solution/sending me down a rabbit hole
Yes, I misread the problem...sorry about that.
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