# 15.1 | Critical Points: Local Extrema and Saddle Points

A function of two variables has a local maximum at $$(a,b)$$ if $$f(x,y) \leq f(a,b)$$ when $$(x,y)$$ is near $$(a,b)$$. The number $$f(a,b)$$ is called a local maximum value. If $$f(x,y) \geq f(a,b)$$ when $$(x,y)$$ is near $$(a,b)$$, then $$f$$ has a local minimum at $$(a,b)$$ and $$f(a,b)$$ is a local minimum value.
Fermat’s Theorem for Functions of Two Variables: If $$f$$ has a local maximum or minimum at $$(a,b)$$ and the first-order partial derivatives of $$f$$ exist there, then $$f_x(a,b) = 0$$ and $$f_y(a,b) = 0$$
Suppose the second partial derivatives of $$f$$ are continuous on a disk with center $$(a,b)$$, and suppose that $$(a,b)$$ is a critical point. Let
\begin{aligned} D = D(a,b) & = f_{xx}(a,b)f_{yy}(a,b) – [f_{xy}(a,b)]^2 \\ & = \left| \begin{array}{c c} f_{xx}(a,b) & f_{xy}(a,b) \\ f_{yx}(a,b) & f_{yy}(a,b) \\ \end{array} \right| \end{aligned}

• If $$D>0$$ and $$f_{xx}(a,b)>0$$, then $$f(a,b)$$ is a local minimum.
• If $$D>0$$ and $$f_{xx}(a,b)<0$$, then $$f(a,b)$$ is a local maximum.
• If $$D<0$$, then $$f(a,b)$$ is a saddle point. That is, $$f(a,b)$$ is not a local maximum or minimum.
• If $$D=0$$, then the test is inconclusive.
Extreme Value Theorem for Functions of Two Variables: If $$f$$ is continuous on a closed, bounded set $$D$$ in $$\mathbb{R}^3$$, then $$f$$ attains an absolute maximum value $$f(x_1,y_1)$$ and an absolute minimum value $$f(x_2, y_2)$$ at some points $$(x_1,y_1)$$ and $$(x_2, y_2)$$ in $$D$$.

Find the local maximum and minimums values and saddle point(s) of
$$f(x,y) = x^2 +xy+y^2+y$$

• Find all the critical points of $$f$$, by finding solutions to the system
$$\begin{cases} f_x(x,y) = 2x +y = 0\\ f_y(x,y) = x + 2y +1 = 0 \end{cases}$$
Solving the first equation for $$y$$ and substituting into the second equation, the system reduces to
$$-3x +1 = 0 \Rightarrow x = \frac{1}{3}$$
If $$x = 1/3$$ then $$y = -2/3$$. Check that this solution satisfies the original system. That is, check that $$f_x(1/3,-2/3) = 0$$ and $$f_y(1/3,-2/3) = 0$$.
• Find $$D(1/3,-2/3)$$ and $$f_{xx}(1/3,-2/3)$$:
$$D(-1/3,-2/3) = \left. \left| \begin{array}{ c c} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{array} \right| \right|_{\left(\frac{1}{3},-\frac{2}{3}\right)}$$
The second partials are
\begin{aligned} f_{xx}(x,y) & = 2 \\ f_{yy}(x,y) & = 2 \\ f_{xy}(x,y) & = 1 \end{aligned}
So that
$$D(-1/3,-2/3) = 4-1 = 3 >0$$

Since $$D(1/3,-2/3)>0$$ and $$f_{xx}(1/3,-2/3)>0$$, the point $$(1/3,-2/3)$$ is a local minimum with minimum value $$f(1/3,-2/3) =-1/3$$

Find the local maximum and minimums values and saddle point(s) of
$$f(x,y) = (x^2+y^2)e^{y^2-x^2}$$

• Find all the critical points of $$f$$, by finding solutions to the system
$$\begin{cases} f_x(x,y) = -2 x e^{y^2-x^2} \left(x^2+y^2-1\right) = 0\\ f_y(x,y) = 2 y e^{y^2-x^2} \left(x^2+y^2+1\right) = 0 \end{cases}$$
Notice that the second equation is satisfied if and only if $$y = 0$$. Substituting this value of $$y$$ into the first equation reduces the system to
$$-2xe^{y^2-x^2}(x^2-1) = 0$$
That is, $$x = 0$$ or $$x= \pm 1$$. So there are three critical points:

• $$(0,0)$$
• $$(1,0)$$
• $$(-1,0)$$
• Find $$D(a,b)$$ and $$f_{xx}(a,b)$$ for each of the critical points:
The second partials are
\begin{aligned} f_{xx}(x,y) & = 2 e^{y^2-x^2} \left(2 x^4+2 x^2 y^2-5 x^2-y^2+1\right) \\ f_{yy}(x,y) & = 2 e^{y^2-x^2} \left(2 x^2 y^2+x^2+2 y^4+5 y^2+1\right) \\ f_{xy}(x,y) & = -4 x y e^{y^2-x^2} \left(x^2+y^2\right) \end{aligned}

• $$D(0,0) = 4$$ and $$f_{xx}(0,0) = 2$$
• $$D(1,0) = -\frac{16}{e^2}$$ and $$f_{xx}(1,0) = -\frac{4}{e}$$
• $$D(-1,0) = -\frac{16}{e^2}$$ and $$f_{xx}(-1,0) = -\frac{4}{e}$$

Since $$D(0,0)>0$$ and $$f_{xx}(0,0)>0$$, the point $$(0,0)$$ is a local minimum with minimum value $$f(0,0) =0$$. The other critical points are saddle points since $$D<0$$ for these points.