...be maximized by making

**R**as large as possible, subject to the constraint that**L**cannot be negative. So,**L**... the maximum, which means the track is circular, with circumference 1 mile...2 Answers · Science & Mathematics · 30/09/2009

...perimeter = 2L + 2π r = 600 ---> L + π r = 300 L = length of straight track > = 200 r = radius of semicircular tracks ... = area = πr² + 2rL L + π r = 300 ==> 200 + π r = 300 r = 100/π...

3 Answers · Science & Mathematics · 18/03/2008

...portion (147.16 ≥ L ≥ 73.08) and then calculate radius of inside of running track ( r =(200- L )/π) Once you've decided on L and r , it is fairly...

1 Answers · Science & Mathematics · 30/10/2009

if your track is circular then it would be 6pi if you go from the inside of 3 to the inside of 5 because 1.5 from 3 to 4 and 1.5 from 4 to 5 so 3 2pi(

**r**) is circumference so 2(3)pi 6pi1 Answers · Science & Mathematics · 15/02/2008

the perimeter is 500 = 2

**L**+ 2 pi**r**and the width of the track is 2r (**L**is the length of the rectangle) with this constraint**L**...2 Answers · Science & Mathematics · 01/02/2010

400 = 2.54 + 2 pi

**R**+ 0.6pi 397.46 = 2 pi**R**+ 0.6 pi 397.46 = 2 pi**R**+ 1.885 395.58 = 2 pi**R**395.58 = 6.28318**R R**= 395.58 / 6.28318**R**= 62.95856 M2 Answers · Science & Mathematics · 10/04/2008

lim{n-->inf}(1+(

**r**/n))^n = lim{n-->inf}e^[nln(1+(**r**/n))] = lim{n-->inf}e^[ln(1+(**r**/n))/(1/n)] = lim{x-->0}e^[ln(1+(rx))/(x)], where x = 1/n = lim{x-->0}e^[**r**/(1+rx)], using**L**'Hopital's rule = e^**r**3 Answers · Science & Mathematics · 26/11/2010

Let the center of the left curve be

**L**, the center of the right curve be**R**. And let the point of intersection of both rails be X . LX...2 Answers · Science & Mathematics · 12/09/2016

A = 2

**r L**+ pi**r**^2 is to be maximized P = 2L + 2 pi**r**= 1500...2 A'(**r**) = 1500 - 4 pi**r**+ 2 pi**r**= 1500 - 2 pi**r**= 0**r**= 1500/(2 pi)**r**= 750/pi so**L**= 0 and there is only the joined semi-circles at the end, no...1 Answers · Science & Mathematics · 01/02/2013

... ( L ), so L = 200 - π r Area of rectangular area inside track = d * L A = 2 r * (200 - π r ) A = 2 (200r - π r²) Find A', set...

1 Answers · Science & Mathematics · 09/11/2009