Maximum Volume of a rectangular box in ellipsoid

Question

This is the problem I am working on:

Find the maximum volume of a rectangular box that can be inscribed in the ellipsoid:

$x^2/25 + y^2/4 + z^2/49 = 1$

with sides parallel to the coordinate axis

I know the Volume equation is going to be $V = 8xyz$. Using Lagrange:

$\nabla V = \lambda\nabla g = \langle8yz, 8xz, 8xy\rangle = \lambda\langle2x/25, 2y/4, 2z/49\rangle$

Solving for y:

$x = 25y/4, z = 49y/4$

Plugging that into $g$, I get $y= .453$, and then $x = 2.831, z = 5.548$, for a $V = 56.9$

This is the wrong answer. Can someone walk me through this and tell me where I went wrong? Thank you in advance!

Answer 1

Hint:

Since $\displaystyle 8yz=\lambda\cdot\frac{2x}{25}, \;\; 8xz=\lambda\cdot\frac{2y}{4},\;\;8xy=\lambda\cdot\frac{2z}{49}$,

multiplying by x in the 1st equation, by y in the 2nd equation, and by z in the 3rd gives

$\hspace{.4 in}\displaystyle\frac{x^2}{25}=\frac{y^2}{4}=\frac{z^2}{49}$.

Now you can solve for x and z in terms of y, say, and then substitute back into the constraint.