# 4.7 Maxima/minima problems  (Page 5/10)

 Page 5 / 10

If the boundary of the set $D$ is a more complicated curve defined by a function $g\left(x,y\right)=c$ for some constant $c,$ and the first-order partial derivatives of $g$ exist, then the method of Lagrange multipliers can prove useful for determining the extrema of $f$ on the boundary. The method of Lagrange multipliers is introduced in Lagrange Multipliers .

## Finding absolute extrema

Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:

1. $f\left(x,y\right)={x}^{2}-2xy+4{y}^{2}-4x-2y+24$ on the domain defined by $0\le x\le 4$ and $0\le y\le 2$
2. $g\left(x,y\right)={x}^{2}+{y}^{2}+4x-6y$ on the domain defined by ${x}^{2}+{y}^{2}\le 16$
1. Using the problem-solving strategy, step $1$ involves finding the critical points of $f$ on its domain. Therefore, we first calculate ${f}_{x}\left(x,y\right)$ and ${f}_{y}\left(x,y\right),$ then set them each equal to zero:
$\begin{array}{ccc}\hfill {f}_{x}\left(x,y\right)& =\hfill & 2x-2y-4\hfill \\ \hfill {f}_{y}\left(x,y\right)& =\hfill & -2x+8y-2.\hfill \end{array}$

Setting them equal to zero yields the system of equations
$\begin{array}{ccc}\hfill 2x-2y-4& =\hfill & 0\hfill \\ \hfill -2x+8y-2& =\hfill & 0.\hfill \end{array}$

The solution to this system is $x=3$ and $y=1.$ Therefore $\left(3,1\right)$ is a critical point of $f.$ Calculating $f\left(3,1\right)$ gives $f\left(3,1\right)=17.$
The next step involves finding the extrema of $f$ on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph: Graph of the domain of the function f ( x , y ) = x 2 − 2 x y + 4 y 2 − 4 x − 2 y + 24 .
${L}_{1}$ is the line segment connecting $\left(0,0\right)$ and $\left(4,0\right),$ and it can be parameterized by the equations $x\left(t\right)=t,y\left(t\right)=0$ for $0\le t\le 4.$ Define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right).$ This gives $g\left(t\right)={t}^{2}-4t+24.$ Differentiating g leads to ${g}^{\prime }\left(t\right)=2t-4.$ Therefore, $g$ has a critical value at $t=2,$ which corresponds to the point $\left(2,0\right).$ Calculating $f\left(2,0\right)$ gives the z- value $20.$
${L}_{2}$ is the line segment connecting $\left(4,0\right)$ and $\left(4,2\right),$ and it can be parameterized by the equations $x\left(t\right)=4,y\left(t\right)=t$ for $0\le t\le 2.$ Again, define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right).$ This gives $g\left(t\right)=4{t}^{2}-10t+24.$ Then, ${g}^{\prime }\left(t\right)=8t-10.$ $g$ has a critical value at $t=\frac{5}{4},$ which corresponds to the point $\left(0,\frac{5}{4}\right).$ Calculating $f\left(0,\frac{5}{4}\right)$ gives the z- value $27.75.$
${L}_{3}$ is the line segment connecting $\left(0,2\right)$ and $\left(4,2\right),$ and it can be parameterized by the equations $x\left(t\right)=t,y\left(t\right)=2$ for $0\le t\le 4.$ Again, define $g\left(t\right)=f\left(x\left(t\right),y\left(t\right)\right).$ This gives $g\left(t\right)={t}^{2}-8t+36.$ The critical value  corresponds to the point $\left(4,2\right).$ So, calculating $f\left(4,2\right)$ gives the z- value $20.$
${L}_{4}$ is the line segment connecting $\left(0,0\right)$ and $\left(0,2\right),$ and it can be parameterized by the equations $x\left(t\right)=0,y\left(t\right)=t$ for $0\le t\le 2.$ This time, $g\left(t\right)=4{t}^{2}-2t+24$ and the critical value $t=\frac{1}{4}$ correspond to the point $\left(0,\frac{1}{4}\right).$ Calculating $f\left(0,\frac{1}{4}\right)$ gives the z- value $23.75.$
We also need to find the values of $f\left(x,y\right)$ at the corners of its domain. These corners are located at $\left(0,0\right),\left(4,0\right),\left(4,2\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(0,2\right)\text{:}$
$\begin{array}{}\\ \hfill f\left(0,0\right)& =\hfill & {\left(0\right)}^{2}-2\left(0\right)\left(0\right)+4{\left(0\right)}^{2}-4\left(0\right)-2\left(0\right)+24\hfill & =\hfill & 24\hfill \\ \hfill f\left(4,0\right)& =\hfill & {\left(4\right)}^{2}-2\left(4\right)\left(0\right)+4{\left(0\right)}^{2}-4\left(4\right)-2\left(0\right)+24\hfill & =\hfill & 24\hfill \\ \hfill f\left(4,2\right)& =\hfill & {\left(4\right)}^{2}-2\left(4\right)\left(2\right)+4{\left(2\right)}^{2}-4\left(4\right)-2\left(2\right)+24\hfill & =\hfill & 20\hfill \\ \hfill f\left(0,2\right)& =\hfill & {\left(0\right)}^{2}-2\left(0\right)\left(2\right)+4{\left(2\right)}^{2}-4\left(0\right)-2\left(2\right)+24\hfill & =\hfill & 36.\hfill \end{array}$

The absolute maximum value is $36,$ which occurs at $\left(0,2\right),$ and the global minimum value is $20,$ which occurs at both $\left(4,2\right)$ and $\left(2,0\right)$ as shown in the following figure. The function f ( x , y ) has two global minima and one global maximum over its domain.
2. Using the problem-solving strategy, step $1$ involves finding the critical points of $g$ on its domain. Therefore, we first calculate ${g}_{x}\left(x,y\right)$ and ${g}_{y}\left(x,y\right),$ then set them each equal to zero:
$\begin{array}{ccc}\hfill {g}_{x}\left(x,y\right)& =\hfill & 2x+4\hfill \\ \hfill {g}_{y}\left(x,y\right)& =\hfill & 2y-6.\hfill \end{array}$

Setting them equal to zero yields the system of equations
$\begin{array}{ccc}\hfill 2x+4& =\hfill & 0\hfill \\ \hfill 2y-6& =\hfill & 0.\hfill \end{array}$

The solution to this system is $x=-2$ and $y=3.$ Therefore, $\left(-2,3\right)$ is a critical point of $g.$ Calculating $g\left(-2,3\right),$ we get
$g\left(-2,3\right)={\left(-2\right)}^{2}+{3}^{2}+4\left(-2\right)-6\left(3\right)=4+9-8-18=-13.$

The next step involves finding the extrema of g on the boundary of its domain. The boundary of its domain consists of a circle of radius $4$ centered at the origin as shown in the following graph. Graph of the domain of the function g ( x , y ) = x 2 + y 2 + 4 x − 6 y .
The boundary of the domain of $g$ can be parameterized using the functions $x\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,y\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t$ for $0\le t\le 2\pi .$ Define $h\left(t\right)=g\left(x\left(t\right),y\left(t\right)\right)\text{:}$
$\begin{array}{cc}\hfill h\left(t\right)& =g\left(x\left(t\right),y\left(t\right)\right)\hfill \\ & ={\left(4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\right)}^{2}+{\left(4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\right)}^{2}+4\left(4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\right)-6\left(4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\right)\hfill \\ & =16{\text{cos}}^{2}\phantom{\rule{0.1em}{0ex}}t+16{\text{sin}}^{2}\phantom{\rule{0.1em}{0ex}}t+16\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t-24\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\hfill \\ & =16+16\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t-24\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t.\hfill \end{array}$

Setting ${h}^{\prime }\left(t\right)=0$ leads to
$\begin{array}{ccc}\hfill -16\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t-24\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t& =\hfill & 0\hfill \\ \hfill -16\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t& =\hfill & 24\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\hfill \\ \hfill \frac{-16\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t}{-16\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t}& =\hfill & \frac{24\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t}{-16\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t}\hfill \\ \hfill \text{tan}\phantom{\rule{0.2em}{0ex}}t& =\hfill & -\frac{4}{3}.\hfill \end{array}$

$\begin{array}{ccc}\hfill -16\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t-24\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t& =\hfill & 0\hfill \\ \hfill -16\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t& =\hfill & 24\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\hfill \\ \hfill \frac{-16\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t}{-16\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t}& =\hfill & \frac{24\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t}{-16\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t}\hfill \\ \hfill \text{tan}\phantom{\rule{0.2em}{0ex}}t& =\hfill & -\frac{3}{2}.\hfill \end{array}$

This equation has two solutions over the interval $0\le t\le 2\pi .$ One is $t=\pi -\text{arctan}\left(\frac{3}{2}\right)$ and the other is $t=2\pi -\text{arctan}\left(\frac{3}{2}\right).$ For the first angle,
$\begin{array}{ccc}\hfill \text{sin}\phantom{\rule{0.2em}{0ex}}t& =\hfill & \text{sin}\left(\pi -\text{arctan}\left(\frac{3}{2}\right)\right)=\text{sin}\left(\text{arctan}\left(\frac{3}{2}\right)\right)=\frac{3\sqrt{13}}{13}\hfill \\ \hfill \text{cos}\phantom{\rule{0.2em}{0ex}}t& =\hfill & \text{cos}\left(\pi -\text{arctan}\left(\frac{3}{2}\right)\right)=\text{−}\text{cos}\left(\text{arctan}\left(\frac{3}{2}\right)\right)=-\frac{2\sqrt{13}}{13}.\hfill \end{array}$

Therefore, $x\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t=-\frac{8\sqrt{13}}{13}$ and $y\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t=\frac{12\sqrt{13}}{13},$ so $\left(-\frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}\right)$ is a critical point on the boundary and
$\begin{array}{cc}\hfill g\left(-\frac{8\sqrt{13}}{13},\frac{12\sqrt{13}}{13}\right)& ={\left(-\frac{8\sqrt{13}}{13}\right)}^{2}+{\left(\frac{12\sqrt{13}}{13}\right)}^{2}+4\left(-\frac{8\sqrt{13}}{13}\right)-6\left(\frac{12\sqrt{13}}{13}\right)\hfill \\ & =\frac{144}{13}+\frac{64}{13}-\frac{32\sqrt{13}}{13}-\frac{72\sqrt{13}}{13}\hfill \\ & =\frac{208-104\sqrt{13}}{13}\approx -12.844.\hfill \end{array}$

For the second angle,
$\begin{array}{ccc}\hfill \text{sin}\phantom{\rule{0.2em}{0ex}}t& =\hfill & \text{sin}\left(2\pi -\text{arctan}\left(\frac{3}{2}\right)\right)=\text{−}\text{sin}\left(\text{arctan}\left(\frac{3}{2}\right)\right)=-\frac{3\sqrt{13}}{13}\hfill \\ \hfill \text{cos}\phantom{\rule{0.2em}{0ex}}t& =\hfill & \text{cos}\left(2\pi -\text{arctan}\left(\frac{3}{2}\right)\right)=\text{cos}\left(\text{arctan}\left(\frac{3}{2}\right)\right)=\frac{2\sqrt{13}}{13}.\hfill \end{array}$

Therefore, $x\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t=\frac{8\sqrt{13}}{13}$ and $y\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t=-\frac{12\sqrt{13}}{13},$ so $\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)$ is a critical point on the boundary and
$\begin{array}{cc}\hfill g\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right)& ={\left(\frac{8\sqrt{13}}{13}\right)}^{2}+{\left(-\frac{12\sqrt{13}}{13}\right)}^{2}+4\left(\frac{8\sqrt{13}}{13}\right)-6\left(-\frac{12\sqrt{13}}{13}\right)\hfill \\ & =\frac{144}{13}+\frac{64}{13}+\frac{32\sqrt{13}}{13}+\frac{72\sqrt{13}}{13}\hfill \\ & =\frac{208+104\sqrt{13}}{13}\approx 44.844.\hfill \end{array}$

The absolute minimum of g is $-13,$ which is attained at the point $\left(-2,3\right),$ which is an interior point of D . The absolute maximum of g is approximately equal to 44.844, which is attained at the boundary point $\left(\frac{8\sqrt{13}}{13},-\frac{12\sqrt{13}}{13}\right).$ These are the absolute extrema of g on D as shown in the following figure. The function f ( x , y ) has a local minimum and a local maximum.

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