
CalculusOptimization
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 $\displaystyle x^2 + y^2 = 4^2$ so $\displaystyle y^2 = 4^2 + x^2$ the $\displaystyle y= sqrroot 16x^2$ and therfore $\displaystyle h=4+y$ so $\displaystyle h=4+ sqrroot 16x^2$
Am I going in the right direction?
Thanks!!

Well, then I plugged h into teh volume formula, so I had:
$\displaystyle V=(1/3)pi r^2 (4+ \sqrt{16x^2})$
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

$\displaystyle V = \frac{\pi}{3} x^2(y+4)$
$\displaystyle V = \frac{\pi}{3}(x^2y+4x^2)$
$\displaystyle \frac{dV}{dx} = \frac{\pi}{3}(x^2 \frac{dy}{dx} + 2xy + 8x)$
setting $\displaystyle \frac{dV}{dx} = 0$ ...
$\displaystyle x \frac{dy}{dx} + 2y + 8 = 0$
since $\displaystyle \frac{dy}{dx} = \frac{x}{y}$ ...
$\displaystyle \frac{x^2}{y} + 2y + 8 = 0$
$\displaystyle x^2 + 2y^2 + 8y = 0$
$\displaystyle (y^2  16) + 2y^2 + 8y = 0$
$\displaystyle 3y^2 + 8y  16 = 0$
$\displaystyle (y+4)(3y4) = 0$
$\displaystyle y = \frac{4}{3}$
$\displaystyle x = \sqrt{16  \frac{16}{9}} = \frac{8\sqrt{2}}{3}$
$\displaystyle V_{max} = \frac{2048\pi}{81}$

now that I think about it, this may be the easier way to go ...
$\displaystyle V = \frac{\pi}{3}x^2(y+4)$
$\displaystyle V = \frac{\pi}{3}(16  y^2)(y+4)$
$\displaystyle V = \frac{\pi}{3}(64 + 16y  4y^2  y^3)$
$\displaystyle \frac{dV}{dy} = \frac{\pi}{3}(16  8y  3y^2)$
$\displaystyle \frac{\pi}{3}(16  8y  3y^2) = 0$
$\displaystyle \frac{\pi}{3}(4+y)(4  3y) = 0$
$\displaystyle y = \frac{4}{3}$ ... and so on.

hmmm...ok, so when I get $\displaystyle y= 4/3$ (since we cant use the 4, cause that just wont work) I plug y back into the $\displaystyle V=(pi/3) x^2 (4+y) $ so I get $\displaystyle (pi/3)x^2(16/3)$ ??
This problem...sheew, im not liking it ..grrrr