Friday

August 29, 2014

August 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

**Related Questions**

Calculus - SOS...help!!! please email me at 969e221 at g mail. com here is the ...

geometry - An isosceles trapezoid with that bases 16 in and 34 in are inscribed ...

calculus - A rectangle is inscribed in an isosceles triangle. If the sides of ...

calculus - find the dimensions of the rectangle of maximum area that can be ...

Math - Find the Maximum area for the given perimeter of a rectangle. State the ...

calculus - a trapezoid is a quadrilateral with (only) two sides (called bases) ...

Please help with a trapezoid - circle problem - A circle with radius 3 is ...

calculus - A rectangle is to be inscribed under the arch of the curve y=4cos(.5x...

Calculus - A rectangle is inscribed with its base on the x -axis and its upper ...

Calculus - A rectangle is inscribed with its base on the x axis and its upper ...