Question

Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Answer 1

Using Lagrange multipliers, we have the Lagrangian


`L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)`

with partial derivatives (set equal to 0)


`L_x=1+2\lambda x=0\implies x=-\frac1{2\lambda}`

`L_y=1+2\lambda y=0\implies y=-\frac1{2\lambda}`

`L_z=-1+2\lambda z=0\implies z=\frac1{2\lambda}`

`L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81`

Substituting the first three equations into the fourth allows us to solve for
`\lambda`:


`x^2+y^2+z^2=\frac1{4\lambda^2}+\frac1{4\lambda^2}+\frac1{4\lambda^2}=81\implies\lambda=\pm\frac1{6\sqrt3}`

For each possible value of
`\lambda`, we get two corresponding critical points at
`(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)`.

At these points, respectively, we get a maximum value of
`f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3` and a minimum value of
`f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3`.