Question

Find the maximum value and minimum value of f(x,y,z)=xyz subject to x+y+z=1 x≥0; y≥0; z≥0

Answer 1

We are asked for the extreme values of $f$ subject to the constraint $g(x,y,z)=x+y+z=1.$

Using Lagrange multipliers, we solve the equations $\nabla f(x,y)=\lambda\nabla g(x, y)$ and $g(x,y)=1,$

which can be written as

$$f_x=\lambda g_x, f_y=\lambda g_y,f_z=\lambda g_z, x+y+z=1$$

$$yz=\lambda, xz=\lambda,xy=\lambda, x+y+z=1$$

$$yz=xz=xy$$

$$yz=xz=>z=0\ or\ x=y$$

If $z=0$

$$xy=0, x+y=1$$

We have $x=0,y=1, z=0$ or $x=1,y=0, z=0.$

If $z\not=0,$ then $x=y.$

If $x=y=0,$ then $z=1.$

If $x=y\not=0,$ then $z=x=y=1/3.$

$$f(0,1,0)=f(1,0,0)=f(0,0,1)=0$$

$$f(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})=\dfrac{1}{3}(\dfrac{1}{3})(\dfrac{1}{3})=\dfrac{1}{27}$$

Therefore the maximum value of $f$ on the plane $x+y+z=1$ is

$$f(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})=\dfrac{1}{27},$$

and the minimum value is

$$f(0,1,0)=f(1,0,0)=f(0,0,1)=0$$