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max and min value 3

Petrus

Well-known member
Feb 21, 2013
739
Calculate highest and lowest value of the function
\(\displaystyle f(x,y)\frac{6}{x^2+y^2+1}+3xy\) in the range
\(\displaystyle \frac{1}{3}≤x^2+y^2≤2\)
Progress
\(\displaystyle f_x(x,y)=3y-\frac{12x}{(x^2+y^2+1)^2}\)
\(\displaystyle f_y(x,y)=3x-\frac{12y}{(x^2+y^2+1)^2}\)
Now when I have derivate respect to x or y I shall find critical point.
\(\displaystyle 3y-\frac{12x}{(x^2+y^2+1)^2}=0\)
\(\displaystyle 3x-\frac{12y}{(x^2+y^2+1)^2}=0\)

Well for the numerator:
\(\displaystyle 3x-12y=0<=>x=4y\)
\(\displaystyle 3y-12x=0<=>3y-48y=0 <=>y=0\) and that means also \(\displaystyle x=0\)
Denominator:
\(\displaystyle 1-(x^2+y^2+1)^2=0 <=>x=\sqrt{+-1-y^2-1^2}\) (I will admit it did take me a while notice I had 1 minus the function :p)
well to input that x into the function is the part I am unsure too handle with.
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's go back to where you have taken the first partials and equated to zero. What you want to do is combine the terms on the left by getting a common denominator, and then look at the numerators, as the denominators have no real roots. You should find 5 critical points, but one of them is not in or on the bounding region.
 
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Petrus

Well-known member
Feb 21, 2013
739
\(\displaystyle f_y(x,y)=\frac{3x(x^2+y^2+1)-12y}{(x^2+y^2+1)}\)
Numerirator:
\(\displaystyle 3x(x^2+y^2+1)-12y=0\) that means \(\displaystyle 3x-12y=0<=>x=4y\)
\(\displaystyle (x^2+y^2+1)-12y=0 <=> x=\sqrt{\sqrt{12y}-y^2-1}\)
Denominator:
\(\displaystyle (x^2+y^2+1)^2=0 <=> x=\sqrt{-y^2-1}\) and that one we shall ignore, is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You should get after simplification:

\(\displaystyle \frac{y(x^2+y^2+1)^2-4x}{(x^2+y^2+1)^2}=0\)

\(\displaystyle \frac{x(x^2+y^2+1)^2-4y}{(x^2+y^2+1)^2}=0\)

We may ignore the denominators as they have no real roots, i.e., the sum of the squares of two real numbers can never be -1. So this reduces the system to:

\(\displaystyle y(x^2+y^2+1)^2-4x=0\)

\(\displaystyle x(x^2+y^2+1)^2-4y=0\)

Can we solve both equations for the same quantity, and then equate those expressions to get a relationship between $x$ and $y$?
 

Petrus

Well-known member
Feb 21, 2013
739
You should get after simplification:

\(\displaystyle \frac{y(x^2+y^2+1)^2-4x}{(x^2+y^2+1)^2}=0\)

\(\displaystyle \frac{x(x^2+y^2+1)^2-4y}{(x^2+y^2+1)^2}=0\)

We may ignore the denominators as they have no real roots, i.e., the sum of the squares of two real numbers can never be -1. So this reduces the system to:

\(\displaystyle y(x^2+y^2+1)^2-4x=0\)

\(\displaystyle x(x^2+y^2+1)^2-4y=0\)

Can we solve both equations for the same quantity, and then equate those expressions to get a relationship between $x$ and $y$?
Can you give me a tips
 

chisigma

Well-known member
Feb 13, 2012
1,704
Calculate highest and lowest value of the function
\(\displaystyle f(x,y)\frac{6}{x^2+y^2+1}+3xy\) in the range
\(\displaystyle \frac{1}{3}≤x^2+y^2≤2\)
Progress
\(\displaystyle f_x(x,y)=3y-\frac{12x}{(x^2+y^2+1)^2}\)
\(\displaystyle f_y(x,y)=3x-\frac{12y}{(x^2+y^2+1)^2}\)
Now when I have derivate respect to x or y I shall find critical point.
\(\displaystyle 3y-\frac{12x}{(x^2+y^2+1)^2}=0\)
\(\displaystyle 3x-\frac{12y}{(x^2+y^2+1)^2}=0\)

Well for the numerator:
\(\displaystyle 3x-12y=0<=>x=4y\)
\(\displaystyle 3y-12x=0<=>3y-48y=0 <=>y=0\) and that means also \(\displaystyle x=0\)
Denominator:
\(\displaystyle 1-(x^2+y^2+1)^2=0 <=>x=\sqrt{+-1-y^2-1^2}\) (I will admit it did take me a while notice I had 1 minus the function :p)
well to input that x into the function is the part I am unsure too handle with.
The problem is greatly simplified if You devide the function by 3 and the convert it in polar coordinates setting $\displaystyle x= \rho\ \cos \theta$ and $y = \rho\ \sin \theta$ so that You have to maximize/minimize the function...

$\displaystyle f(\rho, \theta)= \frac{2}{1+\rho^{2}} + \rho^{2}\ \sin \theta\ \cos \theta$ (1)

... with the condition...

$\displaystyle \frac{1}{3} \le \rho^{2} \le 2$ (2)

First we compute the partial derivatives...

$\displaystyle f_{\rho}(\rho, \theta)= \frac{4\ \rho}{(1+\rho^{2})^{2}}+ 2\ \rho\ \sin \theta\ \cos \theta$

$\displaystyle f_{\theta} (\rho,\theta)= \rho^{2}\ (\cos^{2} \theta- \sin^{2} \theta)$ (3)

... and we observe that $f_{\theta}(*,*)$ vanishes for $\displaystyle \sin \theta = \pm \cos \theta = \pm \frac{1}{\sqrt{2}}$, so that we arrive at the two equation in $\rho$...

$\displaystyle \frac{4}{(1+\rho^{2})^{2}} - 1 =0$ (4)

$\displaystyle \frac{4}{(1+\rho^{2})^{2}} + 1 =0$ (5)

The (5) has no real solutions and the only real positive solution of (4) is $\displaystyle \rho=1$, intenal os the annulus $\displaystyle \frac{1}{\sqrt{3}} \le \rho \le \sqrt{2}$ so that the point of minima or maxima are for $\displaystyle \rho=1$ and $\displaystyle \theta = \frac{\pi}{4},\ \frac{3}{4}\ \pi,\ \frac{5}{4}\ \pi,\ \frac{7}{4}\ \pi$. Further detail are left to You...

Kind regards

$\chi$ $\sigma$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would solve both for $(x^2+y^2+1)^2$, and then equate them.
 

Petrus

Well-known member
Feb 21, 2013
739
We get that \(\displaystyle x=y\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Close...we actually get \(\displaystyle x^2=y^2\)...
 

Petrus

Well-known member
Feb 21, 2013
739
Close...we actually get \(\displaystyle x^2=y^2\)...
Yeah I mean \(\displaystyle +-x=+-y\)
edit: progress will come soon
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would simply use:

\(\displaystyle y=\pm x\)

Now, investigate both cases, and you will find 3 critical points (in my first post in this topic, I originally said there would be 3, then edited my post to say 5, but I was initially correct although not for the right reason).

One case will return only 1 critical point, and the other will return 3, but the two cases share a point, so the total should be 3.
 

Petrus

Well-known member
Feb 21, 2013
739
I would simply use:

\(\displaystyle y=\pm x\)

Now, investigate both cases, and you will find 3 critical points (in my first post in this topic, I originally said there would be 3, then edited my post to say 5, but I was initially correct although not for the right reason).

One case will return only 1 critical point, and the other will return 3, but the two cases share a point, so the total should be 3.
Here is my progress \(\displaystyle y(y^2+y^2+1)^2+4y=0\) so we got \(\displaystyle y+4y=0 <=> y=0\) and then the hard part \(\displaystyle (2y^2+1)^2+4y=0 <=>4y^4+4y^2+1+4y=0\) and if we factor out y \(\displaystyle y(4y^3+4y+4)+1=0 \) and we got then \(\displaystyle y=-1\) and got now problem solve that is left
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is my progress \(\displaystyle y(y^2+y^2+1)^2+4y=0\) so we got \(\displaystyle y+4y=0 <=> y=0\)...
This is incorrect...you want to factor, and then use the zero-factor property and equate both factors to zero and solve for the variable.

\(\displaystyle y(y^2+y^2+1)^2+4y=0\)

Notice both terms have $y$ as a factor...

What you have posted is the case \(\displaystyle x=-y\), and so after this, you need to look at the case \(\displaystyle x=y\).
 

Petrus

Well-known member
Feb 21, 2013
739
This is incorrect...you want to factor, and then use the zero-factor property and equate both factors to zero and solve for the variable.

\(\displaystyle y(y^2+y^2+1)^2+4y=0\)

Notice both terms have $y$ as a factor...

What you have posted is the case \(\displaystyle x=-y\), and so after this, you need to look at the case \(\displaystyle x=y\).
Ok so I get for x and y \(\displaystyle +-\sqrt{\frac{3}{4}}\), \(\displaystyle +-\sqrt{\frac{2}{4}}\) and \(\displaystyle 0\)
(I got only 0 as crit point in that above and rest complex, got rest from \(\displaystyle x=y\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I agree with all except \(\displaystyle x=y=\pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}\)

Can you show us the process by which you obtained these points?
 

Petrus

Well-known member
Feb 21, 2013
739
I agree with all except \(\displaystyle x=y=\pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}\)

Can you show us the process by which you obtained these points?
\(\displaystyle 4y^4+4y^2=3\) so I factour out 4y^2 and get \(\displaystyle 4y^2(y^2+1)=3MATH] I did write wrong point... What do I do wrong?\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Where did you get \(\displaystyle 4y^4+4y^2=3\) ?

edit: Okay, now I see you divided through by $y$, keeping $y=0$ in mind, and obtained:

\(\displaystyle 4y^4+4y^2-3=0\)

So, you would factor to get:

\(\displaystyle (2y^2+3)(2y^2-1)=0\)

Here is the method I used to get the same roots:

In using $x=y$, you should obtain:

\(\displaystyle y(y^2+y^2+1)^2-4y=0\)

\(\displaystyle y\left((2y^2+1)^2-2^2 \right)=0\)

\(\displaystyle y(2y^2+3)(2y^2-1)=0\)

hence:

\(\displaystyle x=y=0,\,\pm\frac{1}{\sqrt{2}}\)
 
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Petrus

Well-known member
Feb 21, 2013
739
Where did you get \(\displaystyle 4y^4+4y^2=3\) ?

edit: Okay, now I see you divided through by $y$, keeping $y=0$ in mind, and obtained:

\(\displaystyle 4y^4+4y^2-3=0\)

So, you would factor to get:

\(\displaystyle (2y^2+3)(2y^2-1)=0\)

Here is the method I used to get the same roots:

In using $x=y$, you should obtain:

\(\displaystyle y(y^2+y^2+1)^2-4y=0\)

\(\displaystyle y\left((2y^2+1)^2-2^2 \right)=0\)

\(\displaystyle y(2y^2+3)(2y^2-1)=0\)

hence:

\(\displaystyle x=y=0,\,\pm\frac{1}{\sqrt{2}}\)
Now that I got my critical point how shall I work with my end point? Shall I use the trick that the one early reply use subsitate on that range so we got \(\displaystyle u=\frac{1}{3}\) and \(\displaystyle u=2\) and put it on orginal function and look for highest and lowest value?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Which of the 3 critical points is outside of the annular bounding region?

After that, you could use the suggestion made by chisigma, or you could use Lagrange multipliers with two cases, or use substitution to get a function in one variable with endpoints.
 

Petrus

Well-known member
Feb 21, 2013
739
Which of the 3 critical points is outside of the annular bounding region?

After that, you could use the suggestion made by chisigma, or you could use Lagrange multipliers with two cases, or use substitution to get a function in one variable with endpoints.
zero cause its less then 1/3
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
zero cause its less then 1/3
Correct, the origin is not on the annulus.

Now, what critical points do you find on the two boundaries?
 

Petrus

Well-known member
Feb 21, 2013
739
Can I subsitute that \(\displaystyle y^2=2-x^2\) or?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, just be mindful that you will have two cases to consider when substituting for $y$, i.e., the positive and negative roots.
 

Petrus

Well-known member
Feb 21, 2013
739
Yes, just be mindful that you will have two cases to consider when substituting for $y$, i.e., the positive and negative roots.
I get the roots \(\displaystyle +-\sqrt(2)\) and one we alredy got \(\displaystyle +-\frac{1}{\sqrt{2}}\) are they correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No what I mean is we have:

\(\displaystyle y^2=2-x^2\) hence:

\(\displaystyle y=\pm\sqrt{2-x^2}\)

and so you would consider the cases:

i) \(\displaystyle f(x)=2+3x\sqrt{2-x^2}\)

ii) \(\displaystyle f(x)=2-3x\sqrt{2-x^2}\)