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November 1st, 2012, 10:39 AM   #1
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Joined: Jun 2011

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a few problems I'm stuck on

Hello,

I attached four jpg files to this post, it is screenshots of problems I had on a quiz last week. I could really use some help understanding how to work them. Some of them I answered correctly because of solutions I found online but could not really understand them I have a test Monday so I'll be doing this all weekend most likely.

Thanks.
Attached Images
 math1.jpg (87.4 KB, 62 views) math2.jpg (101.0 KB, 62 views) math3.jpg (39.5 KB, 62 views)

 November 2nd, 2012, 02:45 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: a few problems I'm stuck on I'll work the first one: The perimeter of the enclosed area is: $13=x+2y+\frac{\pi}{2}x$ The area is: $A=xy+\frac{\pi}{8}x^2$ From the perimeter, we find: $y=\frac{13-x-\frac{\pi}{2}x}{2}$ Hence: $A(x)=x$$\frac{13-x-\frac{\pi}{2}x}{2}$$+\frac{\pi}{8}x^2=\frac{13}{2} x-\frac{4+\pi}{8}x^2$ Equating the derivative to zero, we find: $A'(x)=\frac{13}{2}-\frac{4+\pi}{4}x=0$ $x=\frac{26}{4+\pi}$
 November 2nd, 2012, 09:50 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: a few problems I'm stuck on Here is the last one. The area of the rectangle will be: $A=bh=(2x)(6-x^2)=12x-2x^3$ where $0 $A'(x)=12-6x^2=6$$2-x^2$$=0$ $x=\sqrt{2}$ Hence: $A_{\text{max}}=A$$\sqrt{2}$$=8\sqrt{2}$
 November 3rd, 2012, 12:18 PM #4 Senior Member   Joined: Jun 2011 Posts: 154 Thanks: 0 Re: a few problems I'm stuck on Thanks

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