1
$\begingroup$

Find extrema of $f(x,y,z)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$ subject to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ by reducing variables and then using the Single Variable Method or by using Geometry instead of using the Lagrange Method.

Just to be sure and to get intuition I used the Lagrange method making the constraint $g(x,y,z)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1=0$ and solving $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ from which I got the critical value of $(x^*,y^*,z^*)=(3,3,3)$ which is a saddle point.

Then I tried to reduce variables because I can't think of any geometric argument for this problem. Using the restriction I got $z=\frac{xy}{xy-x-y}$, substituting that in the restriction I got $y=\frac{x}{x-1}$ but when I tried to substitute that into z I get $z=\frac{(x-1)x^2}{(x-1)(x^2-x^2+x-x)}$ but the second part of the denominator is zero. Can someone help?

$\endgroup$
2
$\begingroup$

Do the following substitution,

$$x_1=\frac1x$$ $$y_1=\frac1y$$ $$z_1=\frac1z$$

Then the problem is transformed to, $$f(x_1,y_1,z_1)=x_1^2+y_1^2+z_1^2$$

s.t. $x_1+y_1+z_1=1$

The constraint is a plane passing (0,0,1), (0,1,0), and (1,0,0). The objective function is a sphere with radius r. So the minimum r is when the sphere tangents to the plane or the distance from the origin to the plane that gives you $(\frac13,\frac13,\frac13)$. Thus the maximum point is (3,3,3).

$\endgroup$
1
  • $\begingroup$ Looks good apart from one thing: the extremal point is a (global) minimum point not a maximum point. $\endgroup$ – Winther May 31 '16 at 20:06

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.