# 4.7 Maxima/minima problems  (Page 2/10)

 Page 2 / 10

Find the critical point of the function $f\left(x,y\right)=x3+2xy-2x-4y.$

$\left(2,-5\right)$

The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.

## Definition

Let $z=f\left(x,y\right)$ be a function of two variables that is defined and continuous on an open set containing the point $\left({x}_{0},{y}_{0}\right).$ Then f has a local maximum at $\left({x}_{0},{y}_{0}\right)$ if

$f\left({x}_{0},{y}_{0}\right)\ge f\left(x,y\right)$

for all points $\left(x,y\right)$ within some disk centered at $\left({x}_{0},{y}_{0}\right).$ The number $f\left({x}_{0},{y}_{0}\right)$ is called a local maximum value . If the preceding inequality holds for every point $\left(x,y\right)$ in the domain of $f,$ then $f$ has a global maximum (also called an absolute maximum ) at $\left({x}_{0},{y}_{0}\right).$

The function $f$ has a local minimum at $\left({x}_{0},{y}_{0}\right)$ if

$f\left({x}_{0},{y}_{0}\right)\le f\left(x,y\right)$

for all points $\left(x,y\right)$ within some disk centered at $\left({x}_{0},{y}_{0}\right).$ The number $f\left({x}_{0},{y}_{0}\right)$ is called a local minimum value . If the preceding inequality holds for every point $\left(x,y\right)$ in the domain of $f,$ then $f$ has a global minimum (also called an absolute minimum ) at $\left({x}_{0},{y}_{0}\right).$

If $f\left({x}_{0},{y}_{0}\right)$ is either a local maximum or local minimum value, then it is called a local extremum (see the following figure).

In Maxima and Minima , we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

## Fermat’s theorem for functions of two variables

Let $z=f\left(x,y\right)$ be a function of two variables that is defined and continuous on an open set containing the point $\left({x}_{0},{y}_{0}\right).$ Suppose ${f}_{x}$ and ${f}_{y}$ each exists at $\left({x}_{0},{y}_{0}\right).$ If $f$ has a local extremum at $\left({x}_{0},{y}_{0}\right),$ then $\left({x}_{0},{y}_{0}\right)$ is a critical point of $f.$

## Second derivative test

Consider the function $f\left(x\right)={x}^{3}.$ This function has a critical point at $x=0,$ since $f\prime \left(0\right)=3{\left(0\right)}^{2}=0.$ However, $f$ does not have an extreme value at $x=0.$ Therefore, the existence of a critical value at $x={x}_{0}$ does not guarantee a local extremum at $x={x}_{0}.$ The same is true for a function of two or more variables. One way this can happen is at a saddle point    . An example of a saddle point appears in the following figure.

In this graph, the origin is a saddle point. This is because the first partial derivatives of $f\left(x,y\right)={x}^{2}-{y}^{2}$ are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to $y=0$ is $z={x}^{2}$ (a parabola opening upward), but the vertical trace corresponding to $x=0$ is $z=\text{−}{y}^{2}$ (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.

## Definition

Given the function $z=f\left(x,y\right),$ the point $\left({x}_{0},{y}_{0},f\left({x}_{0},{y}_{0}\right)\right)$ is a saddle point if both ${f}_{0}\left({x}_{0},{y}_{0}\right)=0$ and ${f}_{y}\left({x}_{0},{y}_{0}\right)=0,$ but $f$ does not have a local extremum at $\left({x}_{0},{y}_{0}\right).$

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
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salma
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. 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
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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.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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