find extrema of $2-\left(z-\sqrt{x^2+y^2}\right)^2+\left(z-\sqrt{x^2+y^2}\right)^3$

Question

$$f(x,y,z)=2-\left(z-\sqrt{x^2+y^2}\right)^2+\left(z-\sqrt{x^2+y^2}\right)^3$$

Find maximum and minimum of the function.

Answer 1

Hint: differentiate $f$ wrt $x$, $y$ and $z$ and set $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=0$.

You'll get a system of (three) simultaneous equations in $x$, $y$ and $z$.

Solve these for $(x,y,z)$ and you're done.

...

Reminder: to partially differentiate $f$ wrt $x$ (i.e. to find $\frac{\partial f}{\partial x}$), just differentiate as usual, but treat $y$ and $z$ as constants.

Answer 2

I suggest you to write it in cylindrical coordinates, i.e. $$f(r,\theta,z)=2-\left(z-r\right)^2+\left(z-r\right)^3\ .$$ Now you can take the gradient in cylindrical coordinates (http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates) $$(\frac{\partial f}{\partial r},\frac{1}{r}\frac{\partial f}{\partial \theta},\frac{\partial f}{\partial z})$$ i.e. $$(2(z-r)-3(z-r)^2,0,-2(z-r)+3(z-r)^2)$$ setting it to $0$ you get $$z-r=0\qquad z-r=\frac{2}{3}\ .$$ These define two surfaces of critical points. On the first surface you have the maximum (i.e. 2) and on the second the minimun (i.e. $\frac{26}{27}$).

Answer 3

The extrema of

$$ f(x,y,z) = 2 - \left( z - \sqrt{x^2+y^2} \right)^2 + \left( z- \sqrt{x^2+y^2} \right)^3 $$

Put

$$ \zeta = z - \sqrt{x^2+y^2} $$

so we find the extrema of

$$ f(\zeta) = 2 - \zeta^2 + \zeta^3 $$

which is solving

$$ - 2 \zeta + 3 \zeta^2 = 0$$

Thus

$$\zeta = 0 \textrm{ or } \zeta = \frac{2}{3}$$

As we can write $x^2 + y^2 = \rho^2$, we can write

$$z = \zeta + \rho$$

Thus we have the solutions

$$z = \rho$$

and

$$z = \frac{2}{3} + \rho$$