Extreme Values, Lagrange Multiplier and Taylor's Formula for Two Variables

Extreme Values and Saddle Points

Derivative tests for local extreme values

Definition Let $f(x, y)$ be defined on a region $R$ containing the point $(a, b)$. Then

• $f(a, b)$ is a local maximum value of $f$ if $f(a, b) \geq f(x, y)$ for all domain points $(x, y)$ in an open disk centered at $(a, b)$.

• $f(a, b)$ is a local minimum value of $f$ if $f(a, b) \leq f(x, y)$ for all domain points $(x, y)$ in an open disk centered at $(a, b)$.

Theorem(Theorem 10)

If $f(x, y)$ has a local maximum or minimum value at an interior point $(a, b)$ and if the first partial derivatives exist there, then $f_x(a, b) = 0$ and $f_y(a, b) = 0$.

Critical Points and Saddle Points

Definition (Critical Point)

critical point of $f$: interior point of $f(x, y)$ where both $f_x=f_y=0$ or where one or both of $f_x$ and $f_y$ do not exist.

Definition (Saddle Point)

saddle point $(a, b, f(a,b))$ of the surface $z = f(x, y)$: in every open disk centered at a critical point $(a, b)$ there are domain points $(x, y)$ where $f(x, y) > f(a, b)$ and domain points $(x, y)$ where $f(x, y) < f(a, b)$.

Example

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

Solution:

The domain of $ƒ$ is the entire plane (so there are no boundary points) and the partial derivatives $f_x = -2x$ and $f_y = 2y$ exist everywhere. Therefore, local extrema can occur only at the origin (0, 0) where $f_x = 0$ and $f_y = 0$. Along the positive $x$-axis, however, $f$ has the value $f(x, 0) = -x^2 < 0$; along the positive $y$-axis, $f$ has the value $ƒ(0, y) = y^2 > 0$. Therefore, every open disk in the xy-plane centered at (0, 0) contains points where the function is positive and points where it is negative. The function has a saddle point at the origin and no local extreme values.

Example

Find the local extreme values of $f(x, y) = x^2 + y^2 - 4y + 9$.

Solution:

The domain of ƒ is the entire plane (so there are no boundary points) and the partial derivatives $f_x = 2x$ and $f_y = 2y - 4$ exist everywhere. Therefore, local extreme values can occur only where $$f_x = 2x = 0 , f_y = 2y - 4 = 0.$$ The only possibility is the point $(0, 2)$, where the value of $f$ is 5. Since $f(x, y) =x^2 + (y - 2)^2 + 5$ is never less than 5, we see that the critical point (0, 2) gives a local minimum.

Second derivative test for local extreme values

Suppose that $f(x, y)$ and its first and second partial derivatives are continuous throughout a disk centered at $(a, b)$ and that $f_x(a, b) = f_y(a, b) = 0$. Then

• $f$ has a saddle point at $(a, b)$ if $f_{xx} f_{yy} - f_{xy}^2 < 0$ at $(a, b)$.

• $f$ has a local maximum at $(a, b)$ if $f_{xx} f_{yy} - f_{xy}^2>0$ and $f_{xx}< 0$ at $(a, b)$.

• $f$ has a local minimum at $(a, b)$ if $f_{xx} f_{yy} - f_{xy}^2> 0$ and $f_{xx}> 0$ at $(a, b)$.

• the test is inconclusive at $(a, b)$ if $f_{xx} f_{yy} - f_{xy}^2 = 0$ at $(a, b)$. In this case, we must find another way to determine the behavior of $f$ at $(a, b)$.

Example

Find the local extreme values of $$f(x,y)=xy-x^2-y^2-2x-2y+4.$$

According to second derivative test for local extreme values, $f$ has a local maximum at (-2,-2). The value of $f$ at this point is 8.

Absolute maxima and minima on closed bounded regions

• Check critical points

• Check boundary points

• Compare values.

Example

Find the absolute maximum and minimum values of $$f(x,y)=2+2x+4y-x^2-y^2$$ on the triangular region bounded by the lines $x=0$, $y=0$, and $y=9-x$.

Solution

1. Interior points. For these we have $f_x = 2 - 2x = 0, f_y = 4 - 2y = 0$, yielding the single point $(x, y) = (1, 2)$. The value of $ƒ$ there is $ƒ(1, 2) = 7$.

2. Boundary points. Boundary points. We take the triangle one side at a time:

• On the segment $OA$, $y = 0$. The function $$f(x, y) = f(x, 0) = 2 + 2x - x2$$ may now be regarded as a function of x defined on the closed interval $0 <= x <= 9$. Its extreme values (we know from Chapter 4) may occur at the endpoints $x = 0$ where $f(0, 0) = 2$, $x = 9 where f(9, 0) = 2 + 18 - 81 = -61$ and at the interior points where $f′(x, 0) = 2 - 2x = 0$. The only interior point where $f′(x, 0) = 0$ is $x = 1$, where $f(x, 0) = f(1, 0) = 3$.

• On the segment $OB$, $x = 0$ and $f(x, y) = f(0, y) = 2 + 4y - y2$ the interior point where $f′(0, y) = 0$ occurs at (0, 2), with $f(0, 2) = 6$. So the candidates for this segment are $f(0, 0) = 2, f(0, 9) = -43, f(0, 2) = 6$.

• We have already accounted for the values of $f$ at the endpoints of $AB$, so we need only look at the interior points of the line segment $AB$. With $y = 9 - x$, we have $f(x, y) = 2 + 2x + 4(9 - x) - x^2 - (9 - x)^ 2 = -43 + 16x - 2x^2$ Setting $f′(x, 9 - x) = 16 - 4x = 0$ gives $x = 4$. At this value of $x$, $y = 9 - 4 = 5$ and $f(x, y) = f(4, 5) = -11$.

3. summary We list all the function value candidates: 7, 2, -61, 3, -43, 6, -11. The maximum is 7, which ƒ assumes at (1, 2). The minimum is -61, which ƒ assumes at (9, 0).

Lagrange Multipliers

Constrained maxima and minima

Find point $P(x,y,z)$ on the hyperbolic cylinder $x^2-z^2-1=0$ closest to the origin.

The orthogonal gradient theorem

Suppose that $f(x, y, z)$ is differentiable in a region whose interior contains a smooth curve $$C: \vec{r}(t) = x(t)\vec{i} + y(t)\vec{j} + z(t)\vec{k}.$$ If $P_0$ is a point on $C$ where $f$ has a local maximum or minimum relative to its values on $C$, then $\nabla f$ is orthogonal to $C$ at $P_0$.

The method of Lagrange multipliers

Suppose that $f(x, y, z)$ and $g(x, y, z)$ are differentiable and $\nabla g \neq 0$ when $g(x, y, z) = 0$. To find the local maximum and minimum values of $f$ subject to the constraint $g(x, y, z) = 0$ (if these exist), find the values of $x, y, z,$ and $\lambda$ that simultaneously satisfy the equations $$\nabla f =\lambda \nabla g \qquad {and} \qquad g(x, y, z) = 0.$$

Example

Find the greatest and smallest values of $$f(x,y)=xy$$ takes on the ellipse $${x^2\over 8}+{y^2\over 2}=1.$$

Lagrange multipliers with two constraints

Find the extreme values of a differentiable function $ƒ(x, y, z)$ whose variables are subject to two constraints. If the constraints are $g_1(x, y, z) = 0$ and $g_2(x, y, z) = 0$ and $g_1$ and $g_2$ are differentiable, with $\nabla g_1$ not parallel to $\nabla g_2$, we find the constrained local maxima and minima of $ƒ$ by introducing two Lagrange multipliers $\lambda$ and $\mu$. is, we locate the points $P(x, y, z)$ where $ƒ$ takes on its constrained extreme values by finding the values of $x, y, z, \lambda,$ and $\mu$ that simultaneously satisfy the three equations $$\nabla f =\lambda \nabla g_1+ \mu \nabla g_2,\qquad g_1(x,y,z)=0,\qquad g_2(x,y,z)=0$$

Taylor's Formula for Two Variables