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March 2nd, 2012, 07:03 PM   #1
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Largest area of a rectangle given boundaries

Quote:
 A rectangle is bounded by the x-axis and the semicircle y = sqrt(25 - x^2). What is the largest area that can be formed by a rectangle within these boundaries?
I'm not sure how to get the formula for the area.

It's obviously length x width and one of those is going to be a variable length across the x-axis, and the other side will be the ends of that side to the distance of where the semicircle is and I'm not sure how to find the distance between a point on the x-axis to another point on the semicircle that's vertical. I know they will have the same x-coordinate, but how do I find the variable for the y coordinate?

 March 2nd, 2012, 07:57 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,324 Thanks: 1023 Re: Largest area of a rectangle given boundaries Well, you know the semi-circle has points (5,0), (0,5) and (-5,0) on circumference. So points (a,2a) and (-a,2a) on circumference and (a,0) and (-a,0) on x-axis will be rectangle's coordinates, in other words a square, since a square has max area. Equation of line with points (-a,2a) and (a,2a) is simply y = 2a Circle's is y^2 = 25 - x^2; substitute: (2a)^2 = 25 - a^2 4a^2 + a^2 = 25 a^2 = 5 Since area = (2a)^2, then area = 4a^2 = 4*5 = 20 If you're confused.....graph the semi-circle....
 March 2nd, 2012, 08:19 PM #3 Senior Member   Joined: Jan 2010 Posts: 205 Thanks: 0 Re: Largest area of a rectangle given boundaries Wait, is it a given that a square will always have the highest area given the same set of limited dimensions for any rectangle?
March 2nd, 2012, 09:56 PM   #4
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Re: Largest area of a rectangle given boundaries

Let the base of the rectangle be 2x and the height be y then the area A is:

$A=2xy=2x\sqrt{5^2-x^2}$

Here is a plot of the area function over the domain of the problem $0:

[attachment=0:1sleq1zf]rectinsemicirc.jpg[/attachment:1sleq1zf]

To find the largest area, we need to differentiate the area function and equate the derivative to zero and solve for x to get the critical value.

$\frac{dA}{dx}=2$$x\cdot\frac{-x}{\sqrt{5^2-x^2}}+\sqrt{5^2-x^2}$$=2$$\frac{5^2-2x^2}{\sqrt{5^2-x^2}}$$=0$

Our critical value will come from:

$5^2-2x^2=0$

Taking the positive root, we find:

$x=\frac{5}{\sqrt{2}}$

We shouldn't rely solely upon our graph to ensure there is a maximum at this value, so we may also use the second derivative test:

$\frac{d^2A}{dx^2}=2$$\frac{\sqrt{5^2-x^2}\(-4x$$-$$5^2-2x^2$$$$\frac{-x}{\sqrt{5^2-x^2}}$$}{5^2-x^2}\)=2$$\frac{-4x\(5^2-x^2$$+x$$5^2-2x^2$$}{$$5^2-x^2$$^{\small{\frac{3}{2}}}}\)=$

$\frac{2x$$2x^2-75$$}{$$5^2-x^2$$^{\small{\frac{3}{2}}}}$

We see that on the given domain we have $\frac{d^2A}{dx^2}<0$ so we know our critical value is at a maximum.

Thus, the maximum area is:

$A$$\frac{5}{\sqrt{2}}$$=2$$\frac{5}{\sqrt{2}}$$\sq rt{5^2-$$\frac{5}{\sqrt{2}}$$^2}=\frac{50}{\sqrt{2}}\cdot \sqrt{\frac{2-1}{2}}=25$
Attached Images
 rectinsemicirc.jpg (11.4 KB, 953 views)

 March 2nd, 2012, 10:44 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,324 Thanks: 1023 Re: Largest area of a rectangle given boundaries Yer right (as usual!) Mark; I jumped the gun (as usual!); sorry Daigo: given a certain perimeter, then a square is maximum area; but that's not the case here.
 March 2nd, 2012, 10:49 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Largest area of a rectangle given boundaries If the rectangle had been inscribed within the full circle, then it would have been a square.
 March 2nd, 2012, 10:55 PM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,324 Thanks: 1023 Re: Largest area of a rectangle given boundaries Yes, of course. Well, at least I gave Daigo "the way" to draw a square in a semi-circle
 March 3rd, 2012, 05:47 AM #8 Senior Member   Joined: Jan 2010 Posts: 205 Thanks: 0 Re: Largest area of a rectangle given boundaries Sorry, I'm in an Algebra 2 class so I don't know anything about derivatives and such. Why is the base of the rectangle 2x? Wouldn't it make more sense for the base to be 'x' and the height to be 'y' since Area = L*W can be translated to Area = x*y? I wasn't sure how to sketch graph this as we weren't allowed to use graphing calculators and I was on a time constraint. I thought semicircles were exactly half of a circle. How come the graph looks like a messed up ellipse?
 March 3rd, 2012, 06:11 AM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,324 Thanks: 1023 Re: Largest area of a rectangle given boundaries Daigo, if you can't use derivatives, then are you SURE your original post is correct? Like, if you have to dig a huge hole and have no bulldozer, then you have to use a shovel, right? All yours, Mark

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