Calculate max/min of a 3 variable function, restricted to g(x,y,z)=0

Question

Calculate extrema of $f(x,y,z)=xe^{yz}$ on boundary $3x^2 +y^2 +z^2 =27$

• I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system. $f_x + \lambda g_x$,$f_y + \lambda g_y,f_z + \lambda g_z,g(x,y,z)=0$

$e^{yz}+\lambda (6x)=0$

$xze^{yz}+\lambda (2y)=0$

$xye^{yz}+\lambda (2z)=0$

$3x^2 +y^2 +z^2 -27 = 0$

I tried a lot of combinations but I can't solve this. Can you help me ?

Answer 1

Hint.

Make $\mu = \frac{\lambda}{e^{yz}}$ and then solve

$$ 1+\mu 6 x = 0\\ x z + \mu 2 y = 0\\ xy + \mu 2z = 0\\ 3x^2+y^2+z^2-27=0 $$

giving

$$ \left[ \begin{array}{ccccc} x & y & z & \mu & f \\\ -3 & 0 & 0 & \frac{1}{18} & 0 \\ 3 & 0 & 0 & -\frac{1}{18} & 0 \\ -\frac{1}{\sqrt{3}} & -\sqrt{13} & -\sqrt{13} & \frac{1}{2 \sqrt{3}} & -\frac{13 e}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} & \sqrt{13} & \sqrt{13} & \frac{1}{2 \sqrt{3}} & -\frac{13 e}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} & -\sqrt{13} & -\sqrt{13} & -\frac{1}{2 \sqrt{3}} & \frac{13 e}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} & \sqrt{13} & \sqrt{13} & -\frac{1}{2 \sqrt{3}} & \frac{13 e}{\sqrt{3}} \\ \end{array} \right] $$