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.
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
hi vedant can u help me with some assignments
Solomon
find the 15th term of the geometric sequince whose first is 18 and last term of 387
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=