# Find the critical points of the following functions. Use the Second Derivative Test to determine whether each critical point corresponds to a loal maximum, local minimum, or saddle point. f(x,y)=x^4+2y^2-4xy

Find the critical points of the following functions. Use the Second Derivative Test to determine whether each critical point corresponds to a loal maximum, local minimum, or saddle point.
$f\left(x,y\right)={x}^{4}+2{y}^{2}-4xy$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

lamusesamuset

Let $c=-\left(a,b\right)$ point of the given function. Then we have
$\frac{\partial \left({x}^{4}+2{x}^{2}-4xy\right)}{\partial x}{\mid }_{a,b}=0$ and $\frac{\partial \left({x}^{4}+2{x}^{2}-4xy\right)}{\partial y}{\mid }_{a,b}=0$
which further implies that $\left(4{x}^{3}-4y\right){\mid }_{a,b}=0$ and $\left(4y-4x\right){\mid }_{a,b}=0$ implying that ${a}^{3}-b=0$ and b-a=0 leading to $\left(a,b\right)\equiv \left(0,0\right),\left(1,1\right),\left(-1,-1\right).$
Therefore the required critical points for the given function are (0, 0), (1, 1), and (-1, -1).
Now we calculate the values of $\frac{{\partial }^{2}f}{\partial {x}^{2}},\frac{{\partial }^{2}f}{\partial {y}^{2}}$ and $\frac{{\partial }^{2}f}{\partial x\partial y}$ to be able to use the second derivative test. We get $\frac{{\partial }^{2}f}{\partial {x}^{2}}=\frac{\partial }{\partial x}\left(4{x}^{3}-4y\right),\frac{{\partial }^{2}f}{\partial {y}^{2}}=\frac{\partial }{\partial y}\left(4y-4x\right)$, and $\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}$ mplying that $\frac{{\partial }^{2}f}{\partial {x}^{2}}=12{x}^{2},\frac{{\partial }^{2}f}{\partial {y}^{2}}=4$, and $\frac{{\partial }^{2}f}{\partial x\partial y}=-4$.
Now, we denote $D\left(c\right)=\left(12{x}^{2}{\mid }_{c}\right)\left(4{\mid }_{c}\right)-{\left(\left(-4\right){\mid }_{c}\right)}^{2}$
Now we calculate this for all the above three critical points we found as
$D\left(0,0\right)=\left(12{\left(0\right)}^{24}-{\left(-4\right)}^{2}=-16\right).$
$D\left(1,1\right)=\left(12{\left(1\right)}^{24}-{\left(-4\right)}^{2}=31\right).$
$D\left(-1,-1\right)=\left(12{\left(-1\right)}^{24}-{\left(-4\right)}^{2}=31\right).$
We also have