65.3k views
2 votes
F(x,y) = x^2+y^2+x^2y+6 find the absolute max and min

User Paramjeet
by
8.5k points

1 Answer

2 votes

f(x,y)=x^2+y^2+x^2y+6 has critical points where the partial derivatives simultaneously vanish:


f_x=2x+2xy=2x(1+y)=0\implies x=0\text{ or }y=-1

f_y=2y+x^2=0

x=0\implies 2y=0\implies y=0

y=-1\implies-2+x^2=0\implies x=\pm\sqrt2

So we have three critical points to consider,
(0,0),
(-\sqrt2,-1), and
(\sqrt2,-1).

The function has Hessian


\mathbf H(x,y)=\begin{bmatrix}f_(xx)&f_(xy)\\f_(yx)&f_(yy)\end{bmatrix}=\begin{bmatrix}2+2y&2x\\2x&2\end{bmatrix}

At the critical points, we have


|\mathbf H(0,0)|=\begin{vmatrix}2&0\\0&2\end{vmatrix}=4>0

f_(xx)(0,0)=2>0

which means there is a minimum at (0, 0) of
f(0,0)=6;


|\mathbf H(-\sqrt2,-1)|=\begin{vmatrix}0&amp;-2\sqrt2\\-2\sqrt2&amp;2\end{vmatrix}=-8<0

which means
(-\sqrt2,-1) is a saddle point; and


|\mathbf H(\sqrt2,-1)|=\begin{vmatrix}0&amp;2\sqrt2\\2\sqrt2&amp;2\end{vmatrix}=-8<0

which means
(\sqrt2,-1) is also a saddle point.
User Tunji
by
7.7k points

No related questions found