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November 29, 2014

November 29, 2014

Posted by **calculus** on Monday, October 26, 2009 at 6:27pm.

- Calculus - maximum problem -
**Reiny**, Monday, October 26, 2009 at 7:33pmI will assume you want the base to be on the x-axis.

Let's make the base as long as possible, that is, from (-pi/2,0) to (pi/2,0)

Let the other point of contact be (x,y)

Area of trap = (1/2)(pi + 2x)(y)

= (1/2)(pi+2x)cosx = (1/2)picosx + xcosx

d(Area)/dx = (-1/2)pi(sinx) + x(-sinx) + cosx

= 0 for a max/min of Area

-pi(sinx) - 2xsinx + 2cosx = 0

2cosx = sinx(pi + 2x)

2/(pi + 2x) = tanx

I ran this through a primitive Newton's Method program and got

x = .45797

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