Math 344: Calculus III
14.7 Maximum and Minimum Values
14.7.1 Extrema for Function with One Output ¶
If a function $ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pypx}[2][x]{\dfrac{\partial #2}{\partial #1}} \newcommand{\dydx}[2][x]{\dfrac{d #2}{d #1}} \newcommand{\deltax}[2][x]{\frac{\Delta #2}{\Delta #1}} f:\mathbb{R}^n\rightarrow\mathbb{R} $ has only one output then naturally we want to know if the output has any maximum or minimum values. For example, the paraboloid $z = x^2 + y^2$ has an absolute minimum at the origin of the graph.
Figure 1
Definition ¶
Local Extrema
A function of two variables has a local minimum at $(a,b)$ if there is a disk $D$ of radius $r\gt 0$ centered at the point $(a,b)$ in the domain such that $f(x,y)\ge f(a,b)$ at every point $(x,y)\in D$.A function of two variables has a local maximum at $(a,b)$ if there is a disk $D$ of radius $r\gt 0$ centered at the point $(a,b)$ in the domain such that $f(x,y)\le f(a,b)$ at every point $(x,y)\in D$.
For any disk of radius $r>0$ centered at $(0,0)$ the value of $f(x,y)\ge 0 = f(0,0)$ in our paraboloid. Hence $(0,0)$ is a local minimum for our function $f$.
Definition ¶
Absolute Extrema
A function of two variables has an absolute minimum at $(a,b)$ if $f(x,y)\ge f(a,b)$ for every element $(x,y)$ in the domain.A function of two variables has an absolute maximum at $(a,b)$ if $f(x,y)\le f(a,b)$ for every element $(x,y)$ in the domain.
Our paraboloid has an absolute minimum at the point $(0,0)$. A extremum or extreme point is a minimum or maximum.
Theorem ¶
If a function of two variables $f$ has a local extremum at point $(a,b)$ and both first order partial derivatives $f_1$ and $f_2$ exist at $(a,b)$, then $f_1(a,b) = f_2(a,b)=0$.
Proof ¶
If $f:D\subset\mathbb{R}^2\rightarrow\mathbb{R}$, then define $D_x$ to be the intersection of our domain $D$ with the $x$-axis, and $g:D_x\rightarrow\mathbb{R}$ by $g(x)=f(x,b)$. Then $g$ has an extremum at $x=a$ and $g'(a)=f_x(a,b)$. Since $g$ is differentiable at $x=a$ and $g$ has an extreme value at $x=a$, $g'(a)=0$ from Calculus I. Hence $f_x(a,b) = g'(a) = 0$. Similarly $f_y(a,b)=0$.
If we use our values for $f_x(a,b)$ and $f_y(a,b)$ in the equation of the tangent plane, we have that the equation of the tangent plane is
$$
z = z_0 + f_x(a,b)(x-a) + f_y(a,b)(y-b) = z_0
$$
This means that the tangent plane is parallel to the $xy$-plane. Geometrically this means that if such a function has an extreme value and a tangent plane at a point $(a,b)$, the tangent plane is horizontal at that point.
Definition ¶
A function of two variables for which $f_1(a,b) = f_2(a,b) = 0$ or one of these partial derivatives does not exist is called a critical point or stationary point .
Just as in function of one variable, not all critical points are extrema. A critical point of a function of two variable may be a maximum, minimum or neither.
Example 14.7.1 ¶
Consider $f(x,y) = x^4 + y^4 - 2x^2 + 4y^2 + 1$. Setting the two first order partial derivatives to zero yields
$$
\begin{align*}
f_x(x,y) = 4x^3 - 4x &= 0 &\qquad f_y(x,y) = 4y^3 + 8y &= 0\\
\\
4x(x^2-1) &= 0 &\qquad 4y(y^2+2) &= 0 \\
\\
4x(x-1)(x+1) &= 0 &\qquad 4y(y^2+2) &= 0 \\
\\
x &= 0,1,-1 &\qquad y &= 0 \\
\end{align*}
$$
Thus $(a,b)=(0,0)$, $(1,0)$ and $(-1,0)$ are critical points of $f$ with
$$
\begin{align*}
f(0,0) &= 1 \\
f(1,0) &= 0 \\
f(-1,0) &= 0
\end{align*}
$$
Completing the square in the definition of $f$ gives us
$$
\begin{align*}
f(x,y) &= x^4 + y^4 - 2x^2 + 4y^2 + 1 \\
&= x^4 + y^4 - 2x^2 + 1 - 1 + y^4 + 4y^2 + 4 - 4 + 1 \\
&= \left(x^2 - 1\right)^2 + \left(y^2 + 2\right)^2 - 4
\end{align*}
$$
Though $x^2-1$ may be negative, $\left(x^2-1\right)^2 \ge 0$. Furthermore $y^2\ge 0$, so $y^2+2\ge 2$, and $\left(y^2 + 2\right)^2\ge 4$. Hence
$$
f(x,y) = \left(x^2 - 1\right)^2 + \left(y^2 + 2\right)^2 - 4 \ge 0 + 4 - 4 = 0
$$
Thus $(1,0,0)$ and $(-1,0,0)$ are absolute minimums. Graphically we can see that $(0,0,1)$ is neither a maximum nor a minimum. If we define functions
$$
\begin{align*}
g(x) &= f(x,0), &\ h(y) &= f(0,y) \\
\\
g''(x) &= 12x^2 - 4, &\ h''(y) &= 12y^3 + 8 \\
\\
g''(0) &= -4, &\ h''(0) &= 8
\end{align*}
$$
The level curve $z = f(x,0)$ is concave down, however the level curve $z = f(0,y)$ is concave up. The point $(0,0,1)$ is called a
saddle point
of $f$.
Figure 2
14.7.2 The Second Derivatives Test ¶
Definition ¶
The Second Derivatives Test
Suppose that all four second order partial derivatives of function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ are continuous on a disk containing critical point $(a,b)$. That is $f_x(a,b) = f_y(a,b)=0$. Define $D$ by the determinant
$$ D(a,b) = \begin{vmatrix} f_{xx}(a,b) & f_{xy}(a,b) \\ f_{yx}(a,b) & f_{yy}(a,b) \end{vmatrix} = f_{xx}(a,b)f_{yy}(a,b) - \left[f_{xy}(a,b)\right]^2 $$
(a) If $D\gt 0$ and $f_xx(a,b)>0$ or $f_yy(a,b)>0$, then $f(a,b)$ is a local minimum
(b) If $D\gt 0$ and $f_xx(a,b)<0$ or $f_yy(a,b)<0$, then $f(a,b)$ is a local minimum
(c) If $D\lt 0$ then $f(a,b)$ is a saddle point
(d) If $D=0$, then the test is inconclusive
Example 14.7.2 ¶
Consider the surface $z = \frac{1}{3}x^3 + 2xy − 3x − y^2$
To find the critical points we set the first order partial derivatives to zero,
$$
\begin{align*}
z_x = x^2 + 2y - 3 &= 0 &\ z_y = 2x - 2y &= 0 \\
\\
y^2 + 2y - 3 &= 0 &\ 2x &= 2y \\
\\
(y+3)(y-1) &= 0 &\ x &= y \\
\\
y &= -3,1
\end{align*}
$$
The critical points are $(-3,-3)$ and $(1,1)$.
$$
\begin{align*}
D(x,y) &= \begin{vmatrix} f_{xx}(a,b) & f_{xy}(a,b) \\ f_{yx}(a,b) & f_{yy}(a,b) \end{vmatrix} = \begin{vmatrix} 2x + 2 & 2 \\ 2 & -2 \end{vmatrix} \\
\\
D(-3,-3) &= \begin{vmatrix} -6+2 & 2 \\ 2 & -2 \end{vmatrix} = (-4)(-2) - (2)(2) = 8 - 4 = 4 > 0 \\
f_{xx}(-3,-3) &= -4 < 0 \\
\\
D(1,1) &= \begin{vmatrix} 2 + 2 & 2 \\ 2 & -2 \end{vmatrix} = (4)(-2) - (2)(2) = -8 - 4 = -12 < 0
\end{align*}
$$
Thus $(-3,-3,9)$ is a local maximum and $\left(1,1,-\frac{5}{3}\right)$ is a saddle point.
Figure 3
14.7.3 Absolute Extrema ¶
A
closed
subset of $\mathbb{R}^2$ is a set that contains all of its boundary points. An
open
set is a set that contains none of its boundary points. A
LOT
of sets are neither open nor closed. A
closed ball
or
closed disk
in $\mathbb{R}^2$ of radius $r\gt 0$ centered at the point $(h,k)$ is defined
$$
\overline{D} = \left\{ (x,y)\in\mathbb{R}^2\,:\,(x-h)^2 + (y-k)^2 \le r^2 \right\}
$$
An
open ball
or
open disk
in $\mathbb{R}^2$ of radius $r\gt 0$ centered at the point $(h,k)$ is defined
$$
D = \left\{ (x,y)\in\mathbb{R}^2\,:\,(x-h)^2 + (y-k)^2 \lt r^2 \right\}
$$
The points on the circle, $x^2 + y^2 = r^2$ are
boundary points
of both sets.
A bounded set is a set that is entirely contained inside of an open disk centered at the origin. Some bounded sets may need a disk with a VERY big radius to contain it! The rectangle $[5280,6000]\times[-1,1]$ is far from the origin, but contained in the disk of radius 12,000 centered at the origin.
Theorem ¶
The Extreme Value Theorem for Function of Two Variables
If $f$ is continuous on a closed bounded set $D$ in $\mathbb{R}^2$, 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 $D$.
Finding the absolute extrema of a function on a closed bounded set $D\subset\mathbb{R}^2$ requires several steps.
-
Find the values of $f$ at the critical points that are in the domain $D$. This requires finding the points where the tangent plane is horizontal, or one of the first order partials does not exist in the domain $D$.
-
Find the extrema of the boundary points. This usually requires methods from Calculus I.
-
The largest and smallest values of $f$ in steps one and two are the absolute extreme values.
Example 14.7.3 ¶
Consider the surface $g(x,y) = x^2 + y^2 + 4x − 6y$ on the domain defined by $D = \left\{ (x,y)\,:\,x^2 + y^2 \le 16 \right\}$.
$$
\begin{align*}
g_x = 2x + 4 &= 0 &\ g_y = 2y - 6 &= 0 \\
\\
2x &= -4 &\ 2y &= 6 \\
\\
x &= -2 &\ y &= 3 \\
\end{align*}
$$
The point $(-2,3)$ is the only critical point in the open interior of the disk. On the boundary circle of the disk we can parameterize the curve by $z = \left(4\cos(\theta), 4\sin(\theta)\right)$ on the interval $[0,2\pi]$. Thus the value of our function on the boundary
$$
\begin{align*}
z(\theta) &= 16\cos^2(\theta) + 16\sin^2(\theta) + 16\cos(\theta) - 24\sin(\theta) \\
\\
&= 16 + 16\cos(\theta) - 24\sin(\theta)
\\
z'(\theta) &= -16\sin(\theta) - 24\cos(\theta) = 0 \\
\\
24\cos(\theta) &= -16\sin(\theta) \\
\\
\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)} &= -\frac{24}{16} = -\dfrac{3}{2} \\
\\
\theta &= \tan^{-1}\left(-\frac{3}{2}\right), \tan^{-1}\left(-\frac{3}{2}\right) + \pi
\end{align*}
$$
The values of our function at are approximately
$$
\begin{align*}
f(-2,3) &= -13) \\
f\left(4\cos(\theta), 4\sin(\theta)\right) &\approx 44.84 \\
f\left(-4\cos(\theta),-4\sin(\theta)\right) &\approx -12.84
\end{align*}
$$
Thus the absolute maximum at $\left(4\cos(\theta), 4\sin(\theta)\right)$ and the absolute minimum occurs at $(-2,3)$.
Figure 4
Exercise 14.7.1 ¶
Find the absolute minimum and maximum values of the function
$$ f(x,y) = x^2 + xy + y^2 - 6y $$
on the set
$$ D = \left\{(x,y) : x\in [-3,3],\ y\in [0,5]\right\} $$
Check Your Work
Minimum: $f\left(-2,4\right) = -12$. Maximum: $f\left(3,5\right) = 19$.Video Solution
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