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.
In this morden time nanotechnology used in many field .
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2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
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and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.