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#1 2006-09-15 02:50:43

Jessie8957
Member
Registered: 2006-09-15
Posts: 3

Please Help

I am taking a calculus class and i need help with the following problem:
An athletic field consists of a rectangular region with a semicircular region at each end.  The perimeter will be used for a 440 yard track.  Find the value of ‘x’ for which the area of the rectangular region is as large as possible. I JUST NEED TO KNOW HOW TO GET STARTED AND WHAT FORMULA I NEED TO USE. If anyone could please help

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#2 2006-09-15 03:05:11

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Please Help

what does x refer to? the radius of the semi circular region, edge of rectangular with semi circle? or what?


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#3 2006-09-16 12:52:58

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Re: Please Help

x=????????


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#4 2006-09-16 13:03:06

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Please Help

I believe x may refer to a side of the rectangle.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-09-16 15:41:40

Jessie8957
Member
Registered: 2006-09-15
Posts: 3

Re: Please Help

I think refers to the side of the rectangle and this was a optimiaztion problem were you use a derivitive formula to help you with finding the answer. Anyone know about derivtive formalas?

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#6 2006-09-17 04:16:44

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Please Help

But which side of the rectangle is x?  I'm pretty sure it's not going to matter though.

Perimeter = 440 = 2y + 2*pi*r

2d = r
x = d
2x = r
r = 1/2x

perimeter = 440 = 2y + pi*x

2y = 440 - pi*x
y = 220 - pi*x/2

area of the rectangle = x * y = x (220 - pi*x/2)

So

we need to find the value for x for which f(x) = x(220 - pi*x/2) is at it's max.

Here's where the calculus comes in...  How do we find the max?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#7 2006-09-17 14:01:50

Jessie8957
Member
Registered: 2006-09-15
Posts: 3

Re: Please Help

thanks so much very helpful

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