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!



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.

  • $\begingroup$ Wow. Not sure how I missed that. Thank you! Using that I got the right answer! $\endgroup$ – Kommander Kitten Apr 6 '15 at 0:37
  • $\begingroup$ Great - I'm glad that helped! $\endgroup$ – user84413 Apr 6 '15 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.