Here is another max-min problem.
When you want to maximize or minimize a positive quantity that involves the square root, raise that quantity to the power 2

Question

Here is another max-min problem.
When you want to maximize or minimize a positive quantity that involves the square root, raise that quantity to the power 2

anonymous · Answer

attachment

lastdaywork · Answer

(e)

anonymous · Answer

Yes, can you show your steps.

lastdaywork · Answer

$$R ^{2}=h ^{2}+x ^{2}$$
$$V=\pi x ^{2}h=\pi (R ^{2}-h ^{2})h$$

By dV/dh = 0 ; we get $$R^2=3h^2$$
implies
$$x=R \sqrt{\frac{ 2 }{ 3 }}$$

anonymous · Answer

Here is the solution generated by http://saab.org

lastdaywork · Answer

http://saab.org ^^ Is that your website ??

anonymous · Answer

Yes

ganeshie8 · Answer

Lagrange also looks simple this time :-
$V = \pi x^2h$
$g(x,h) : h^2+x^2-R^2 = 0 $

$Vx =\lambda   g_x  \implies    2 \pi xh = \lambda  2x \implies  \lambda = \pi h $
$Vh =\lambda   g_y  \implies    \pi x^2 = \lambda  2h $

$ \implies  \pi x^2  = \pi h *  2 h  \implies  x^2 = 2h^2 $

$h^2 + x^2 - R^2 = 0$
$x^2/2 +x^2 - R^2 = 0$