# Maximum Volume

• Jul 7th 2009, 06:17 AM
Maximum Volume
Find the maximum volume of a box with faces parallel to the
xy, xz, yz-planes that lies entirely inside the region enclosed by the cones

z = sqrt(x^2 + 2y^2) - 2 and

z = 2 - sqrt(x^2 + 2y^2)

• Jul 7th 2009, 07:20 AM
malaygoel
The enclosed region is symmetric about xy-plane.

So, if z=k is on of the plane, then the other is z=-k.

If we cut the enclosed area by the plane z=k, we will get an ellipse
$\displaystyle x^2+2y^2=(2-k)^2$

or,$\displaystyle \frac{x^2}{(2-k)^2}+\frac{y^2}{0.5(2-k)^2}=1$

The area of the rectangle having largest area(with sides parallel to ellipse axes) is $\displaystyle \sqrt 2 (2-k)^2$

Hence, the volume of box is $\displaystyle 2\sqrt 2 k(2-k)^2$

For maximum volume, value of k is found to be $\displaystyle k=\frac{2}{3}$
• Jul 8th 2009, 05:21 AM
BobP
What method are you supposed to use ?
It looks like a Lagrange Multiplier problem to me.
Choose a point say $\displaystyle (p,q,r)$ on the cone
$\displaystyle z = 2 - \sqrt(x^2 + 2y^2)$
in the first octant, (so $\displaystyle p,q,r$ are all positive ), and form your box based on this point. It will have a volume of $\displaystyle 8pqr$.
Now let
$\displaystyle W = 8pqr - \lambda\{2 - r - \sqrt(p^2 + 2q^2)\}$
and differentiate partially wrt $\displaystyle p,q,r$ and $\displaystyle \lambda$.
Put each partial derivative equal to zero and solve the resulting equations simultaneously.
I get the values $\displaystyle p = 2\sqrt(2)/3, q = r = 2/3$,
in which case the maximum volume is $\displaystyle 64\sqrt(2)/27.$