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March 4th, 2017, 07:20 AM  #1 
Newbie Joined: Mar 2017 From: Marrakech Posts: 2 Thanks: 0  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 $\displaystyle (2x)*(yb)$, 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: $\displaystyle 2x*[\sqrt(r^2x^2)b]$. When I differentiate (and set to zero to maximize), I get: $\displaystyle (2x^2)/\sqrt(r^2x^2)+2\sqrt(r^2x^2)2b=0$ 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. 
March 4th, 2017, 12:35 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
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. 
March 4th, 2017, 01:02 PM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 
derivative set equal to 0 ... $\dfrac{x^2}{\sqrt{r^2x^2}} + \sqrt{r^2x^2}  b = 0$ Quote:
$\dfrac{y^2r^2}{y} + y  b = 0$ $y  \dfrac{r^2}{y} + y  b = 0$ $2y  \dfrac{r^2}{y}  b = 0$ $2y^2  by  r^2 = 0$ $y = \dfrac{b \pm \sqrt{(b)^2  4(2)(r^2)}}{2(2)}$ $y = \dfrac{b \pm \sqrt{b^2+8r^2}}{4}$ 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 $b=5$ $A = 2x(\sqrt{7^2x^2}  5)$ with its max value is shown in the attached graph (note y in the graph represents the max area, not the yvalue for the rectangle) $y = \dfrac{5+\sqrt{5^2+8(7^2)}}{4} \approx 6.355144464$ $x = \sqrt{7^2y^2} \approx 2.934644585$ $A = 2x(y5) \approx 7.953734727$  
March 4th, 2017, 03:57 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
I would let the objective function be: $\displaystyle f(x,y)=xy$ Subject to the constraint: $\displaystyle g(x,y)=x^2+(y+b)^2r^2=0$ $\displaystyle 0<b<r$ 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: $\displaystyle y(y+b)+(y+b)^2r^2=0$ Taking the positive root, we obtain: $\displaystyle y=\frac{\sqrt{b^2+8r^2}3b}{4}$ 
March 5th, 2017, 01:40 AM  #5  
Newbie Joined: Mar 2017 From: Marrakech Posts: 2 Thanks: 0  Quote:
Quote:
I like your solution a lot, making things a lot easier just cutting straight to the chase (to y). Thanks for showing me Quote:
Thanks for a different sort of solution/sending me down a rabbit hole thanks guys  
March 5th, 2017, 06:12 AM  #6  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Quote:
 

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circular, inside, maximization, problem, rectangle, segment 
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