Lagrange Multipliers, calc max volume of box

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SUMMARY

The discussion focuses on using the Method of Lagrange Multipliers to maximize the volume of a rectangular parallelepiped inscribed within the ellipsoid defined by the equation 2x² + 10y² + 5z² = 80. The function to maximize is f(x,y,z) = xyz, with the constraint g(x,y,z) = 2x² + 10y² + 5z² - 80 = 0. The solution involves deriving equations from the gradients and substituting to find the maximum volume, ultimately leading to the coordinates of point P as x = √5y, z = √2y, and y = √(8/3).

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Homework Statement



Point P(x,y,z) lies on the part of the ellipsoid 2x^2 + 10y^2 + 5z^2 = 80 that is in the first octant of space. It is also a vertex of a rectangular parallelpiped each of whose sides are parallel to a coordinate plane. Use Method of LaGrange Multipliers to determine the coordinates of P so that the box has a max volume and calculate the max



Homework Equations


f(x,y,z)=xyz g(x,y,z)=2x^2+10y^2+5z^2=80


The Attempt at a Solution



\nablaf=\nablag\lambda

1.yz=4x\lambda
2.xz=20y\lambda
3.xy=10z\lambda

I multiplied equation 1 by x, 2 by y and 3 by z

4x^2\lambda=20y^2\lambda=10z^2\lambda

I then put x and z in terms of y and put into constraint
4x^2=20y^2 10z^2=20y^2
x=\sqrt{}5 y z=\sqrt{}2 y

g=2(\sqrt{}5y)^2 +10y^2 + 5(\sqrt{}2y)^2=80
solving for y=\sqrt{}(8/3)

I'm not sure if I'm on the right track or if this is way off, if correct do I just do the same proceedure to find x and z?
 
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You are doing fine. You don't need to repeat the procedure for x and z. You already have x=sqrt(5)*y and z=sqrt(2)*y. Once you've got y, you've got everything.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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