3.7 Maxima and Minima

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In this section we determine local maxima and minima of a surface.

Critical Point A critical point for the function $f(x,y)$ is a point $(x_0, y_0)$ in the domain of $f$ for which either $\grad f(x_0, y_0) = \vec 0$ or one of $f_x$ or $f_y$ (or both) is undefined at $(x_0, y_0)$ .

Example 1 Find the critical points of the function $f(x,y) = x^2 + y^2 - 6y$ .
We set the gradient equal to the zero vector, and we solve for $x$ and $y$ : $\grad f(x,y) = \vector {2x, 2y -6} = \vector {0,0}$ which gives $x = 0$ and $y = 3$ . So the only critical point of $f$ is $(0,3)$ .

(Problem 1) Find the critical points of the function $f(x,y) = x+ y - xy$ .

Example 2 Find the critical points of the function $f(x,y) = 2x^2 - y^2 +2xy + 3x + y+ 5$ .
We set the gradient equal to the zero vector, and we solve for $x$ and $y$ : $\grad f(x,y) = \vector {4x+2y + 3, 2x - 2y + 1} = \vector {0,0}$ which gives the $2 \times 2$ linear system: \begin{align*} 4x+2y + 3 &= 0\\ 2x -2y + 1 & = 0 \end{align*}

Adding the equations gives $6x + 4 = 0$ so $x = -2/3$ . Plugging this into either of the original two equations gives $y = -1/6$ . Hence, $f$ has one critical point: $(-2/3, -1/6)$ .

(Problem 2) Find the critical points of the function $f(x,y) = xy - 2x-2y-x^2 - y^2$ .

Example 3 Find the critical points of the function $f(x,y) = x^2y +xy^2 +3x + 4$ .
We set the gradient equal to the zero vector, and we solve for $x$ and $y$ : $\grad f(x,y) = \vector {2xy + y^2 + 3, x^2 + 2xy} = \vector {0,0}$ which gives the $2 \times 2$ system: \begin{align*} 2xy + y^2 +3 &= 0\\ x^2 + 2xy & = 0 \end{align*}

Solving the bottom equation for $y$ gives $y = -\frac {x}{2}$ and substituting this into the top equation give $2x\left (-\frac {x}{2}\right ) + \left (-\frac {x}{2}\right )^2 +3 = 0$ which simplifies to $-x^2 + \frac 14 x^2 + 3 =0$ $\frac 34 x^2 = 3$ $x^2 = 4$ and finally, we have $x=\pm 2$ . Since $y = -x/2$ , when $x=2$ we have $y=-1$ and when $x = -2$ we have $y = 1$ . Thus, the two critical points are $(2,-1) \quad \text {and} \quad (-2, 1)$

(Problem 3) Find the critical points of the function $f(x,y) = 12xy - x^3 - y^3$ .

Example 4 Find the critical points of the function $f(x,y) = x^4 + y^3 + 2x^2y^2 -27y + 1$ .
We set the gradient equal to the zero vector, and we solve for $x$ and $y$ : $\grad f(x,y) = \vector {4x^3 + 4xy^2, 3y^2 + 4x^2y -27} = \vector {0,0}$ which gives the $2 \times 2$ system: \begin{align*} 4x^3 + 4xy^2 &= 0\\ 3y^2 + 4x^2y -27 &= 0 \end{align*}

The first equation can be written as $4x(x^2 + y^2) = 0$ from which we can deduce that $x = 0$ or $(x, y) = (0,0)$ . Letting $x = 0$ in the bottom equation gives $3y^2 - 27 = 0$ from which we can conclude that $y = \pm 3$ . We also see that $(0,0)$ is not a critical point. Hence, the critical points are $(0, 3)$ and $(0, -3)$ .

(Problem 4) Find the critical points of the function $f(x,y) = 2-x^4 + 2x^2 - y^2$ .

Fermat’s Theorem If $f(x,y)$ has a local extreme at the point $(x_0, y_0)$ , then $(x_0, y_0)$ is a critical point for $f(x,y)$ .

To determine the nature of $f(x,y)$ at a critical point, we use the second derivatives of $f$ .

Second Derivatives Test Let $(x_0, y_0)$ be a critical point for the function $f(x,y)$ , and consider the discriminant, $D(x,y) = f_{xx}(x,y) \cdot f_{yy}(x,y) - \left (f_{xy}(x,y)\right )^2$ 1) If $D(x_0, y_0) > 0$ and $f_{xx}(x_0, y_0) > 0$ , then $f(x,y)$ has a local minimum at $(x_0, y_0)$ .
2) If $D(x_0, y_0) > 0$ and $f_{xx}(x_0, y_0)< 0$ , then $f(x,y)$ has a local maximum at $(x_0, y_0)$ .
3) If $D(x_0, y_0) < 0$ , then $(x_0, y_0)$ is a saddle point.
4) If $D(x_0, y_0) = 0$ , then there is no conclusion about $f(x,y)$ at $(x_0, y_0)$ .

In parts 1 and 2 of the Second Derivatives Test, the condition $f_{xx}(x_0, y_0) > 0$ can be replaced with $f_{yy}(x_0, y_0) > 0$ since they both have the same sign when the discriminant is positive.

Example 5 Use the Second Derivatives Test to determine the nature of $f(x,y) = x^2 + y^2 - 6y$ at its critical points.
In Example 1, we found that the only critical point of $f$ is $(0,3)$ . The second derivatives are $f_{xx} = 2,\; f_{yy} = 2 \quad \text {and} \quad f_{xy} = 0$ The discriminant is $D(x,y) = f_{xx} f_{yy} - f_{xy}^2 = 4$ and at the critical point, it is $D(0,3) = 4$ Since $D(0, 3) > 0$ and $f_{xx}(0,3) = 2 > 0$ , the Second Derivatives Test tells us that $f(x,y) = x^2 + y^2 - 6y$ has a local minimum at the critical point $(0,3)$ .

(Problem 5) Use the Second Derivatives Test to determine the nature of $f(x,y) = x+ y - xy$ (from Problem 1) at its critical points.

Example 6 Use the Second Derivatives Test to determine the nature of $f(x,y) = 2x^2 - y^2 +2xy + 3x + y+ 5$ at its critical points.
In Example 2, we found that the only critical point of $f$ is $(-2/3,-1/6)$ . The second derivatives are $f_{xx} = 4,\; f_{yy} = -2 \quad \text {and} \quad f_{xy} = 2$ The discriminant is $D(x,y) = f_{xx} f_{yy} - f_{xy}^2 = -8 -4 = -12$ and at the critical point, it is $D(-2/3,-1/6) = -12$ Since $D(-2/3,-1/6) < 0$ , the Second Derivatives Test tells us that $f(x,y) = 2x^2 - y^2 +2xy + 3x + y+ 5$ has a saddle point at the critical point $(-2/3,-1/6)$ .

(Problem 6) Use the Second Derivatives Test to determine the nature of $f(x,y) = xy - 2x-2y-x^2 - y^2$ (from Problem 2) at its critical points.

Here is a video solution of problem 6:

Example 7 Use the Second Derivatives Test to determine the nature of $f(x,y) = x^2y +xy^2 +3x + 4$ at its critical points.
In Example 3, we found that the critical points of $f$ are $(2,-1)$ and $(-2, 1)$ . The second derivatives are $f_{xx} = 2y,\; f_{yy} = 2x \quad \text {and} \quad f_{xy} = 2x + 2y$ The discriminant is $D(x,y) = f_{xx} f_{yy} - f_{xy}^2 = 4xy - (2x+2y)^2 = 4xy - 4x^2 - 8xy - 4y^2 = -4(x+y)^2$ and at the critical points we have $D(2, -1) = -4 \quad \text {and} \quad D(-2, 1) = -4$ Since $D < 0$ , at both critical points, the Second Derivatives Test tells us that $f(x,y) = x^2y +xy^2 +3x + 4$ has saddle points at both of the critical points $(2,-1)$ and $(-2, 1)$ .

(Problem 7) Use the Second Derivatives Test to determine the nature of $f(x,y) = 12xy - x^3 - y^3$ (from Problem 3) at its critical points.

Here is a video solution of problem 7:

Example 8 Use the Second Derivatives Test to determine the nature of $f(x,y) = x^4 + y^3 + 2x^2y^2 -27y + 1$ at its critical points.
In Example 4, we found that the critical points of $f$ are $(0, 3)$ and $(0, -3)$ . The second derivatives are $f_{xx} = 12x^2 + 8y^2,\; f_{yy} = 6y + 8x^2 \quad \text {and} \quad f_{xy} = 8xy$ The discriminant is $D(x,y) = f_{xx} f_{yy} - f_{xy}^2 = (12x^2 + 8y^2)(6y + 8x^2) - (8xy)^2$ and at the critical points, we have $D(0, 3) = (72)(18) - (0)^2 = 1296 > 0$ and $D(0, -3) = (72)(-18) - (0)^2 = -1296 < 0$ Since $D(0, 3) > 0$ , the Second Derivatives Test tells us that $f$ has a local extreme at $(0, 3)$ . Furthermore, since $f_{xx} (0,3) = 72 > 0$ , it tells us that this local extreme is a local minimum. At the other critical point, since $D(0, -3) < 0$ , the Second Derivatives Test tells us that $f$ has a saddle point at $(0, -3)$ .

(Problem 8) Use the Second Derivatives Test to determine the nature of $f(x,y) = 2-x^4 + 2x^2 - y^2$ (from Problem 4) at its critical points.

Here is a video solution of problem 8:

Use the 3D grapher to inspect the surfaces in the examples and problems near their critical points.

2025-06-05 13:07:12