Find the maximum volume of a box with faces parallel to thexy, 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)

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- July 7th 2009, 06:17 AMJoAdams5000Maximum VolumeFind the maximum volume of a box with faces parallel to thexy, 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)

- July 7th 2009, 07:20 AMmalaygoel
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

or,

The area of the rectangle having largest area(with sides parallel to ellipse axes) is

Hence, the volume of box is

For maximum volume, value of k is found to be - July 8th 2009, 05:21 AMBobP
What method are you supposed to use ?

It looks like a Lagrange Multiplier problem to me.

Choose a point say on the cone

in the first octant, (so are all positive ), and form your box based on this point. It will have a volume of .

Now let

and differentiate partially wrt and .

Put each partial derivative equal to zero and solve the resulting equations simultaneously.

I get the values ,

in which case the maximum volume is