# Calculus-Optimization

• Oct 21st 2008, 05:53 PM
BCHurricane89
Calculus-Optimization
Ok, this problem is killing me, can I get some help?

Find the volume of the largest(maximize) right circular cone that can be inscribed in a sphere of radius 4.
http://i34.tinypic.com/6id7bc.gif

So, so far ive gotten $x^2 + y^2 = 4^2$ so $y^2 = 4^2 + x^2$ the $y= sqrroot 16-x^2$ and therfore $h=4+y$ so $h=4+ sqrroot 16-x^2$

Am I going in the right direction?
Thanks!!
• Oct 21st 2008, 06:34 PM
BCHurricane89
Well, then I plugged h into teh volume formula, so I had:

$V=(1/3)pi r^2 (4+ \sqrt{16-x^2})$

Then I have to find the derivative of it, but that is what is driving me nuts.

this was the deriv I got, but how in the world do I simplify this thing? (My algebra isnt that great im finding out.. :()http://i33.tinypic.com/1251jwo.jpg
• Oct 21st 2008, 06:38 PM
skeeter
$V = \frac{\pi}{3} x^2(y+4)$

$V = \frac{\pi}{3}(x^2y+4x^2)$

$\frac{dV}{dx} = \frac{\pi}{3}(x^2 \frac{dy}{dx} + 2xy + 8x)$

setting $\frac{dV}{dx} = 0$ ...

$x \frac{dy}{dx} + 2y + 8 = 0$

since $\frac{dy}{dx} = -\frac{x}{y}$ ...

$-\frac{x^2}{y} + 2y + 8 = 0$

$-x^2 + 2y^2 + 8y = 0$

$(y^2 - 16) + 2y^2 + 8y = 0$

$3y^2 + 8y - 16 = 0$

$(y+4)(3y-4) = 0$

$y = \frac{4}{3}$

$x = \sqrt{16 - \frac{16}{9}} = \frac{8\sqrt{2}}{3}$

$V_{max} = \frac{2048\pi}{81}$
• Oct 21st 2008, 07:13 PM
skeeter
now that I think about it, this may be the easier way to go ...

$V = \frac{\pi}{3}x^2(y+4)$

$V = \frac{\pi}{3}(16 - y^2)(y+4)$

$V = \frac{\pi}{3}(64 + 16y - 4y^2 - y^3)$

$\frac{dV}{dy} = \frac{\pi}{3}(16 - 8y - 3y^2)$

$\frac{\pi}{3}(16 - 8y - 3y^2) = 0$

$\frac{\pi}{3}(4+y)(4 - 3y) = 0$

$y = \frac{4}{3}$ ... and so on.
• Oct 21st 2008, 07:35 PM
BCHurricane89
hmmm...ok, so when I get $y= 4/3$ (since we cant use the -4, cause that just wont work) I plug y back into the $V=(pi/3) x^2 (4+y)$ so I get $(pi/3)x^2(16/3)$ ??

This problem...sheew, im not liking it ..grrrr