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

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Answer 1 · 2017-08-18T15:07:28

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

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We calculate the partial derivatives of the function

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

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

$f_{x} = 4 x^{3} + 2 z^{2}$ $f_x=4x^3+2z^2$

$f_{z} = 4 x z - 882 z$ $f_z=4xz-882z$

The critical points are when $f_{x} = 4 x^{3} + 2 z^{2} = 0$ $f_x=4x^3+2z^2=0$ , $\Rightarrow$ $=>$ , $x = 0$ $x=0$ and $z = 0$ $z=0$

$f_{z} = 4 x z - 882 z = 0$ $f_z=4xz-882z=0$

$= 2 z (2 x - 441) = 0$ $=2z(2x-441)=0$

$\Rightarrow$ $=>$ , $z = 0$ $z=0$ and $x = \frac{441}{2}$ $x=441/2$

$f_{x x} = 12 x^{2}$ $f_( x x)=12x^2$

$f_{z z} = 4 x - 882$ $f_(z z)=4x-882$

$f_{x z} = 4 z$ $f_(x z)=4z$

$f_{z x} = 4 z$ $f_(zx)=4z$

$D (x, z) = f_{x x} f_{z z} - f_{x z}^{2}$ $D(x,z)=f_( x x)f_(z z)-f_(x z)^2$

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

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

$f_{x} (0, \frac{441}{2}) = 4 \cdot {(\frac{441}{2})}^{3} > 0$ $f_x(0,441/2)=4*(441/2)^3>0$

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

This test is inconclusive

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

What are the values and types of the critical points, if any, of f(x,z)=x4+15z2+2xz2−456z2f(x,z) = x^4 + 15z^2 + 2xz^2 - 456z^2?