# What are the values and types of the critical points, if any, of f(x,z) = x^4 + 15z^2 + 2xz^2 - 456z^2?

Aug 18, 2017

There is only a relative minimum at $= \left(0 , \frac{441}{2}\right)$

#### Explanation:

We calculate the partial derivatives of the function

$f \left(x , z\right) = {x}^{4} + 15 {z}^{2} + 2 x {z}^{2} - 456 {z}^{2}$

$= {x}^{4} + 2 x {z}^{2} - 441 {z}^{2}$

${f}_{x} = 4 {x}^{3} + 2 {z}^{2}$

${f}_{z} = 4 x z - 882 z$

The critical points are when ${f}_{x} = 4 {x}^{3} + 2 {z}^{2} = 0$, $\implies$, $x = 0$ and $z = 0$

${f}_{z} = 4 x z - 882 z = 0$

$= 2 z \left(2 x - 441\right) = 0$

$\implies$, $z = 0$ and $x = \frac{441}{2}$

${f}_{x x} = 12 {x}^{2}$

${f}_{z z} = 4 x - 882$

${f}_{x z} = 4 z$

${f}_{z x} = 4 z$

$D \left(x , z\right) = {f}_{x x} {f}_{z z} - {f}_{x z}^{2}$

$D \left(0 , \frac{441}{2}\right) = 12 {x}^{2} \left(4 x - 882\right) - 16 {z}^{2}$

$= 12 \cdot {\left(\frac{441}{2}\right)}^{2} - 0 > 0$

${f}_{x} \left(0 , \frac{441}{2}\right) = 4 \cdot {\left(\frac{441}{2}\right)}^{3} > 0$

$D \left(0 , 0\right) = 12 {x}^{2} \left(4 x - 882\right) - 16 {z}^{2} = 0$

This test is inconclusive

There is only a relative minimum at $= \left(0 , \frac{441}{2}\right)$