Question

(4) warning: in general checking along the axes only is not enough to determine if a point is a local minimum or maximum. for more on this look up the multivariable second derivative test in the reference book. (a) let $f(x,y)=4x^{2}+8y^{2}+7$. find the critical point(s) of $f(x,y)$, and determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or neither (\saddle point\)

Explanation:

Step1: Find the first - order partial derivatives

First, find $f_x$ and $f_y$.
$f_x=\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(4x^{2}+8y^{2}+7)=8x$
$f_y=\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}(4x^{2}+8y^{2}+7)=16y$

Step2: Find the critical points

Set $f_x = 0$ and $f_y = 0$.
$8x=0\Rightarrow x = 0$
$16y=0\Rightarrow y = 0$
So the critical point is $(0,0)$.

Step3: Find the second - order partial derivatives

$f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=\frac{\partial}{\partial x}(8x)=8$
$f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial}{\partial y}(16y)=16$
$f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=\frac{\partial}{\partial y}(8x)=0$

Step4: Use the second - derivative test

The discriminant $D=f_{xx}(0,0)f_{yy}(0,0)-[f_{xy}(0,0)]^{2}$
Substitute the values: $D=(8)(16)-0^{2}=128$
Since $f_{xx}(0,0)=8>0$ and $D = 128>0$, the function has a local minimum at the point $(0,0)$.

Answer:

The critical point is $(0,0)$ and it is a local minimum.