I am having a bit of trouble setting up a problem:
(1) Find the volume of the largest regtangular box with edges parallel to the axes that can be inscribed in the ellipsoid: 9x^2 + 36y^2 + 4z^2 = 36
I thought it would be simply:
f(x,y,z) = xyz && g(x,y,z) = 9x^2 + 36y^2 + 4z^2 = 36
However, this seems to give me the wrong solution.
Any help would be appreciated.
I agree with your setup except that you need to multiply xyz by a constant to get the volume.
In two dimensions with an ellipse,
max xy subject to 9x^2 + 36y^2 = 36
would produce a point (x,y) on the ellipse in the upper right quadrant. The inscribed rectangle would have corners (x,y), (x,-y), (-x,y) and (-x,-y). The area of the rectangle would be 4xy.
In three dimensions, the constant to multiply xyz by is 8.
Hello, machi4velli!
This is the way I was taught to use Lagrange Multipliers.(1) Find the volume of the largest regtangular box with edges parallel to the axes
that can be inscribed in the ellipsoid: .
The length of the box is , the width , the height .
. . The volume of the box is: . .
We have: . .and .
. . Then: .
Set the four partial derivatives equal to zero, and solve.
[1] .
[2] .
[3] .
[4] .
From [1], we have: . .[5]
From [2], we have: . .[6]
From [3], we have: . .[7]
Equate [5] and [6]: . .[8]
Equate [6] and [7]: . .[9]
Substitute [8] and [9] into [4]: .
. . and we get: .
Substitute into [8]: .
Substitute into [9]: .
. . Hence: .
Therefore, the maximum volume is: . unitsł