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March 2nd, 2012, 07:03 PM  #1  
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  Largest area of a rectangle given boundaries Quote:
It's obviously length x width and one of those is going to be a variable length across the xaxis, 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 xaxis to another point on the semicircle that's vertical. I know they will have the same xcoordinate, 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: 12,077 Thanks: 794  Re: Largest area of a rectangle given boundaries
Well, you know the semicircle 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 xaxis 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 semicircle.... 
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 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  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: Here is a plot of the area function over the domain of the problem : [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. Our critical value will come from: Taking the positive root, we find: 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: We see that on the given domain we have so we know our critical value is at a maximum. Thus, the maximum area is: 
March 2nd, 2012, 10:44 PM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,077 Thanks: 794  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: 12,155 Thanks: 466 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: 12,077 Thanks: 794  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 semicircle 
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: 12,077 Thanks: 794  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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