# How do I determine the maximum and minimum points for this problem using the Lagrange multiplier approach?

Here is the original problem:

Find the extrema of $$f(x,y)=xyz$$ on the unit ball $$xyz$$ on the unit ball $$x^2+y^2+z^2 \le 1$$.

Here is what I got:

$$f_x = yz, f_y = xz, f_z = xy$$

$$g_x=2x, g_y=2y, g_z =2z$$

$$yz = \lambda2x$$

$$xz = \lambda2y$$

$$xy = \lambda2z$$

$$x^2+y^2+z^2 \le 1$$

=$$(yz/2\lambda)^2 + (xz/2\lambda)^2 + (xy/2\lambda)^2 \le 1$$

=$$(yz)^2 + (xz)^2 + (xy)^2 \le 2\lambda$$

=$$(y^2z^2) + (x^2z^2) + (x^2y^2) \le 2\lambda$$

I solved for lambda:

$$xz = 2\lambda(x^2z^2)$$

$$1/2 = 2\lambda$$

$$1/4 = \lambda$$

...then used lambda to solve for x in terms of yz

$$yz = 2x\lambda$$

$$yz= 2x(1/4)$$

$$yz = x/2$$

$$2yz = x$$

... and plugging it into the equation above ($$xy = \lambda2z$$):

$$xy = 2z\lambda$$

$$(2yz)y = 2z(1/4)$$

$$2y^2z = 2z(1/4)$$

= $$y^2 = 1/4$$

=$$y = \pm 1/2$$

Plugging in the rest I got ($$1/2$$) for x and z also. $$1/2$$ seems to work for the above equations but somehow this feels wrong. I tried this before and got $$x= \pm1/8, y=\pm1/8, and z=\pm1/2.$$ What am I doing wrong?

With that said, you get these four relations all as equality relations. Substituting gets $$xz=4\lambda^2 z/x$$ which yields $$x=\pm 2\lambda$$ and same for $$y$$ and $$z$$, assuming none of them are zero (which is easily seen by inspection to not be the case). So each variable is $$\pm 2 \lambda$$, i.e. they are all the same magnitude (the actual value of $$\lambda$$ is not really important). Plug into the constraint to finish.
You made a mistake in your derivation in that you didn't square $$2\lambda$$, resulting in the wrong value of $$\lambda$$.