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:
$f\left(x,y\right)={x}^{2}-2xy+4{y}^{2}-4x-2y+24$ on the domain defined by
$0\le x\le 4$ and
$0\le y\le 2$
$g\left(x,y\right)={x}^{2}+{y}^{2}+4x-6y$ on the domain defined by
${x}^{2}+{y}^{2}\le 16$
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:
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:
${L}_{1}$ 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
$\left(0,0\right),\left(4,0\right),\left(4,2\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(0,2\right)\text{:}$
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.
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:
The solution to this system is
$x=\mathrm{-2}$ and
$y=3.$ Therefore,
$\left(\mathrm{-2},3\right)$ is a critical point of
$g.$ Calculating
$g\left(\mathrm{-2},3\right),$ we get
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.
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
$h\left(t\right)=g\left(x\left(t\right),y\left(t\right)\right)\text{:}$
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,
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
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
The absolute minimum of
g is
$\mathrm{-13},$ which is attained at the point
$\left(\mathrm{-2},3\right),$ 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.
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