# Relative Extrema and the Second Partials Test

• By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## Relative Extrema

Definition. Let $f$ be a function of two variables $x$ and $y.$

The function $f$ has a relative maximum at $\left(x_0,y_0\right)$ if $f(x,y)\leq f\left(x_0,y_0\right)$ for all $(x,y)$ in an open disk containing $\left(x_0, y_0\right).$

The function $f$ has a relative minimum at $\left(x_0,y_0\right)$ if $f(x,y)\geq f\left(x_0,y_0\right)$ for all $(x,y)$ in an open disk containing $\left(x_0, y_0\right).$

Collectively, relative maxima and relative minima are called relative extrema.

Theorem. (Critical Points) If $f$ has a relative extremum at $\left(x_0,y_0\right)$ and partial derivatives $f_x$ and $f_y$ both exist at $\left(x_0,y_0\right),$ then $$f_x \left(x_0, y_0\right) = f_y\left(x_0, y_0\right) = 0.$$

Proof. Let $F(x)=f\left(x,y_0\right).$ Then $F(x)$ must have a relative extremum at $x=x_0,$ so $F’\left(x_0\right)=0,$ which means that $f_x\left(x_0,y_0\right) = 0.$ Similarly, $G(y)=f\left(x_0,y\right)$ has a relative extremum at $y=y_0,$ so $G’\left(y_0\right)=0$ and $f_y\left(x_0,y_0\right)=0.$ Thus, we must have both $f_x\left(x_0,y_0\right)=0$ and $f_y\left(x_0,y_0\right)=0.$

Definition. A critical point of a function $f$ defined on an open set $D$ is a point $\left(x_0,y_0\right)$ in $D$ where either one of the following is true:

(1) $f_x\left(x_0,y_0\right)=f_y\left(x_0,y_0\right)=0$ or

(2) at least one of $f_x$ or $f_y$ does not exist at $\left(x_0,y_0\right).$

Example. Find the critical points for the function $$f(x,y)=x^2+y^2-2x-6y+14.$$

Solution. The first partial derivatives of $f$ are $$f_x(x,y)=2x-2 \qquad \text{and} \qquad f_y(x,y)=2y-6.$$ These partial derivatives are equal to 0 when $x=1$ and $y=3$ so the only critical point is $(1,3).$ By completing the square we find that $$f(x,y)=4+(x-1)^2+(y-3)^2.$$ Since $(x-1)^2\geq 0$ and $(y-3)^2\geq 0,$ we have $f(x,y)\geq 4$ for all values of $x$ and $y.$ Therefore, $f(1,3)=4$ is a local minimum, and in fact it is the absolute minimum of $f.$ This can be confirmed geometrically from the graph of $f$, which is the elliptic paraboloid with vertex $(1,3,4)$ as shown.

Definition. A critical point $P_0\left(x_0,y_0\right)$ is called a saddle point of a function $f$ if every open disk centered at $P_0$ contains points in the domain of $f$ that satisfy $f(x,y)>f\left(x_0,y_0\right)$ as well as points in the domain of $f$ that satisfy $f(x,y)<f\left(x_0,y_0\right).$

The next example illustrates what a saddle point is.

Example. Find the extreme values of $f(x,y)=y^2-x^2.$

Solution. Since $f_x=-2x$ and $f_y=2y,$ the only critical point is $(0,0).$ Notice that for points on the $x$-axis we have $y=0,$ so $f(x,0)=-x^2<0$ (if $x\neq 0.$) However for points on the $y$-axis we have $x=0,$ so $f(0,y)=y^2>0$ (if $y\neq 0$). Thus every disk with center $(0,0)$ contains points where $f$ takes positive values as well as points where $f$ takes negative values. Therefore $f(0,0)=0$ cannot be an extreme value for $f,$ so $f$ has no extreme values.

This example illustrates the fact that a function need not have a maximum or minimum value at a critical point. The graph of $f$ is the hyperbolic paraboloid which has a horizontal tangent plane $(z=0)$ at the origin. You can see that $f(0,0)=0$ is a maximum in the direction of the $x-\text{axis}$ but not in the direction of the $y$-axis. Near the origin the graph has the shape of a saddle as shown.

Theorem. (Second Partials Test) Assume $f$ has a critical point at $P_0(x_0,y_0)$ and assume that $f$ has continuous second order partial derivatives in a disk centered at $\left(x_0,y_0\right).$ If $D:=(f_{x x}f_{y y}-\left(f_{x y}\right)^2)\left(x_0,y_0\right)>0$, then

(1) a relative maximum occurs at $P_0$ if $f_{x x}\left(x_0,y_0\right)<0$

(2) a relative minimum occurs at $P_0$ if $f_{x x}\left(x_0,y_0\right)>0.$

If $D\left(x_0,y_0\right)<0$, then a saddle point occurs at $P_0.$

Example. Find the relative extrema of the function $$f(x,y)=x^4+y^4-4x y+1.$$

Solution. We first locate the critical points $$f_x=4x^3-4y \qquad \text{and} \qquad f_y=4y^3-4x.$$ Setting these partial derivatives to 0, we obtain the equations$x^3-y=0$ and $y^3-x=0.$ To solve these equations we substitute $y=x^3$ from the first equation into the second one. This gives \begin{align*} 0 & =x^9-x =x\left(x^8-1\right) \\ & =x\left(x^4 -1 \right) \left(x^4+1 \right) \\ & = x(x-1)(x+1)\left(x^2+1\right)\left(x^4+1\right) \end{align*} So there are three real roots: $x=0,1 ,-1.$ The three critical points are $(0,0), (1,1),$ and $(-1,-1).$ Next we calculate the second partial derivatives and $D(x,y)$: $$f_{x x}=12x^2, \qquad f_{x y}=-4, \qquad f_{y y}=12y^2.$$ Thus, $$D(x,y)=f_{x x}f_{y y}-\left(f_{x y}\right)^2=144x^2y^2-16.$$ Since $D(0,0)=-16<0,$ it follows that the origin is a saddle point; that is, $f$ has no local extremum at $(0,0).$ Since $D(1,1)=128>0$ and $f_{x x}(1,1)=12>0,$ we see that $f(1,1)=-1$ is a local minimum. Similarly, we have $D(-1,-1)=128>0$ and $f_{x x}(-1,1-1)=12>0,$ so $f(-1,-1)=-1$ is also a local minimum.

## Exercises on Relative Extrema

Exercise. Find the absolute extrema for the function $f(x,y)=x y^2-2xy+3y$ in the triangular domain with vertices $(0,0)$, $(1,0)$, and $(1,1).$

Exercise. Find and classify the relative extrema and the saddle points of $f(x,y)=xy -2x-4y.$

Exercise. Find the critical points of the function
$f(x,y)=x^3+y^2-2xy+7x-8y+2.$

Exercise. Find and classify the relative extrema and the saddle points of $f(x,y)=xy -2x-4y.$

Exercise. Find the absolute extrema for the function $f(x,y)=x y^2-2xy+3y$ in the triangular domain with vertices $(0,0)$, $(1,0)$, and $(1,1).$

Exercise. Find the critical points and classify each as a relative maximum, a relative minimum, or a saddle point.

$(1) \quad f(x,y)=2x^2-4 x y+y^3+2$

$(2) \quad f(x,y)=e^{-x}\sin y$

$(3) \quad f(x,y)=(x-2)^2+(y-3)^4$

$(4) \quad f(x,y)=\left(x^2+2y^2\right)e^{1-x^2-y^2}$

$(5) \quad f(x,y)=x^2+y^2+\frac{32}{x y}$

$(6) \quad f(x,y)=(x-4)\ln (x y)$

$(7) \quad f(x,y)=2x^3+y^3+3x^2-18y^2+81y+5$

$(8) \quad f(x,y)=3x^2+12x+8y^3-12y^2+7$

Exercise. Find the absolute maximum and minimum values for each of the following functions.

$(1) \quad f(x,y)=e^{x^2+2x+y^2}$ on the disk $x^2+2x+y^2\leq 0.$

$(2) \quad f(x,y)=x^2+x y +y^2$ on the disk $x^2+y^2\leq 1.$

$(3) \quad f(x,y)=x y-2x-5y$ on the triangular region $S$ with vertices $(0,0),$ $(7,0),$ and $(7,7).$

$(4) \quad f(x,y)=x^2-4 x y+y^3+4y$ on the square $0\leq x\leq 2,$ $0\leq y\leq 2.$

$(5) \quad f(x,y)=x^2+3y^2-4x+2y-3$ on the square region $S$ with vertices $(0,0),$ $(3,0),$ $(3,-3),$ and $(0,-3).$

Exercise. Find three positive numbers whose sum is 54 and whose product is as large as possible.

Exercise. A wire of length $L$ is cut into three pieces that are bent to form a circle, a square, and a equilateral triangle. How should the cuts be made to minimize the sum of the total area.?

Exercise. A rectangular box with no top is to have a fixed volume. What should its dimensions be if we want to use the least amount of material in its
construction?

Exercise. Let $R$ be the triangular region in the $x y$-plane with vertices $(-1,-2),$ $(-1,2),$ and $(3,2).$ A plate in the shape of $R$ is heated so that the temperature at $(x,y)$ is $$T(x,y)=2x^2-x\text{ }y+y^2-2y+1$$ (in degrees Celsius). At what point in $R$ or on its boundary is $T$ maximized? What are the temperatures?

Exercise. Find the maximum and minimum values for the given function in the given closed region.

$(1) \quad z=8x^2+4y^2+4y+5$, $x^2+y^2\leq 1$

$(2) \quad z=6x^2+y^3+6y^2$, $x^2+y^2\leq 25$

$(3) \quad z=8x^2-24xy+y^2$, $x^2+y^2\leq 25$

Exercise. Find the volume of the largest box that cane inscribed in the ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.$$

Exercise. Assuming that the function $$F= 2x^2+6y^2+45z^2-4xy+6yz+12xz-6y+14$$ has a minimum, find it. Prove that this function has no maximum value.