Question

Max(min) f(x,y,z) = x^2+y^2+z^2
Subject to
X+2y+z=30
2x-y-3z= 10
Check for local second order condition

Answer 1

Solution:

$$x=30-2y-z$$

$$2(30-2y-z)-y-3z=10$$

$$60-4y-2z-y-3z=10$$

$$50-5y-5z=0$$

$$10-y-z=0$$

$$y=10-z$$

$$x=30-2\times (10-z)-z$$

$$x=30-20+2z-z$$

$$x=10+z$$

$$f(x;y;z)\to f(z)$$

$$f(z)=(10+z)^2+(10-z)^2+z^2$$

$$f(z)=100+20z+z^2+100-20z+z^2+z^2$$

$$f(z)=3z^2+200$$

$$f'(z)=6z$$

$$6z=0; z=0$$

With this value, a minimum of function is achieved.

$$x=10; y=10$$

$$f(x;y;z)=10^2+10^2+0^2=200$$

Answer: 200, minimum