Question

Find the extreme values of f subject to both constraints

F (x , y , z ) = z ; x^2 + y^2 = z^2 , x + y + z = 24

Answer 1

The Lagrangian is


`L(x,y,z,\lambda_1,\lambda_2)=z+\lambda_1(x^2+y^2-z^2)+\lambda_2(x+y+z-24)`

with partial derivatives (set equal to zero) yielding


`\begin{cases}L_x=2x\lambda_1+\lambda_2=0\\L_y=2y\lambda_1+\lambda_2=0\\L_z=1-2z\lambda_1+\lambda_2=0\\L_(\lambda_1)=x^2+y^2-z^2=0\\L_(\lambda_2)=x+y+z-24=0\end{cases}`

Subtracting the second equation from the first gives


`(2x\lambda_1+\lambda_2)-(2y\lambda_1+\lambda_2)=2\lambda_1(x-y)=0\implies x=y`

So in the fourth and fifth equations, we have


`\begin{cases}2x^2=z^2\\2x+z=24\end{cases}\implies x=24\pm12\sqrt2,z=-24\mp24\sqrt2`

There are then two critical points for
`f(x,y,z)=z` at
`(24+12\sqrt2,24+12\sqrt2,-24-24\sqrt2)` and
`(24-12\sqrt2,24-12\sqrt2,-24+24\sqrt2)`.

Clearly,
`f(x,y,z)=z` attains a local minimum at the first point of
`-24-24\sqrt2`, and a local maximum at the second point of
`-24+24\sqrt2`.