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max and min value

Petrus

Well-known member
Feb 21, 2013
739
Decide max and min value for
\(\displaystyle f(x,y)=x^2+y^2+2x-2y+2\) where \(\displaystyle x^2+y^2 \leq1\)
progress:
for solving that equation \(\displaystyle f_x(x,y), f_y(x,y)\) we get the point \(\displaystyle (1,-1)\)
if we do parametric \(\displaystyle x= \cos(t) \, y= \sin(y)\) where \(\displaystyle 0\leq t \leq 2\pi\)
if we put those value and simplify and derivate we get
\(\displaystyle -2\sin(t)-2\cos(t)=0 <=> 1=-\tan(t)\)
that means its on second and 4th quadrants
do I take \(\displaystyle -tan^{-1}\) now?
and get now got \(\displaystyle -tan^{-1}(t)\)
so we got \(\displaystyle x=\cos(-\tan^{1}(1)) <=> t=\frac{1}{2}\), \(\displaystyle x=\sin(-\tan^{1}(1)) <=> t=\frac{1}{2}\)
is this correct?
so we got \(\displaystyle (-\frac{1}{2},\frac{1}{2})\) and \(\displaystyle (\frac{1}{2},-\frac{1}{2})\)
Regards,
\(\displaystyle |\pi\rangle\)
 
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HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Decide max and min value for
\(\displaystyle f(x,y)=x^2+y^2+2x-2y+2\) where \(\displaystyle x^2+y^2 \leq1\)
progress:
for solving that equation \(\displaystyle f_x(x,y), f_y(x,y)\) we get the point \(\displaystyle (1,-1)\)
But \(\displaystyle (-1)^2+ 1^2= 2> 1\) so (1, -1) is not in the desired region and is irrelevant.

if we do parametric \(\displaystyle x= \cos(t) \, y= \sin(y)\) where \(\displaystyle 0\leq t \leq 2\pi\)
if we put those value and simplify and derivate we get
\(\displaystyle -2\sin(t)-2\cos(t)=0 <=> 1=-\tan(t)\)
that means its on second and 4th quadrants
"On second and 4th quadrants" includes a lot of territory! \(\displaystyle tan(t)= -1\) when \(\displaystyle t= 3\pi/4\) and \(\displaystyle T= -\pi/4\)

do I take \(\displaystyle -tan^{-1}\) now?
and get now got \(\displaystyle -tan^{-1}(t)\)
so we got \(\displaystyle x=\cos(-\tan^{1}(1)) <=> t=2\),
It's hard to tell what you mean. "\(\displaystyle -tan^{-1}\)" is meaningless without an argument. I take it you mean \(\displaystyle tan^{-1}(-1)\) which gives the values I give above: \(\displaystyle t= 3\pi/4\) and \(\displaystyle t= -\pi/4\)

\(\displaystyle x=\sin(-\tan^{1}(1)) <=> t=2\)
is this correct?

Regards,
\(\displaystyle |\pi\rangle\)
I have no clue where "t= 2" cae from. There are two ways to find "\(\displaystyle \sin(\tan^{-1}(-1))[/tex]. One is to use the value of the inverse tan I gave above: \(\displaystyle \sin(3\pi/4)= \frac{\sqrt{2}}{2}\) and \(\displaystyle sin(-\pi/4)= -\frac{\sqrt{2}}{2}\). The other is to imagine a right triangle formed with vertices (0, 0), (-1, 0), (-1, 1) so that each leg has length 1 and the tangent is -1/1= -1. Then the hypotenuse has length \(\displaystyle \sqrt{2}\) so "opposite side over hypotenuse" is \(\displaystyle \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}\). Then iagine a right triangle formed with vertices (0, 0), (1, 0), (1, -1) so that, again, each leg has length 1 and the tangent is 1/-1= -1. Again the hypotenuse has length \(\displaystyle \sqrt{2}\) so "opposite side over hypotenuse" is \(\displaystyle \frac{-1}{\sqrt{2}}= -\frac{\sqrt{2}}{2}\).

Similarly for cosine.

(And you had started with x= cos(t), y= sin(t) so you have x and y reversed. Because of the symmetry, that won't matter.)\)
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Decide max and min value for
\(\displaystyle f(x,y)=x^2+y^2+2x-2y+2\) where \(\displaystyle x^2+y^2 \leq1\)
progress:
for solving that equation \(\displaystyle f_x(x,y), f_y(x,y)\) we get the point \(\displaystyle (1,-1)\)
Is $f_{x} = 2\ x + 2$ and $f_{y} = 2\ y - 2$, so that You have $f_{x}=f_{y}=0$ in (-1,1). If You pass to the second derivatives You find $f_{x\ x}=2$, $f_{x\ y}=0$, $f_{y\ x} =0$ and $f_{y\ y}=2$, so that the Hessian determinant is H = 4 > 0 and the point (-1,1) is a minimum. That means that elsewhere f(x,y) has a greater value... what does that suggest to You?...

Kind regards

$\chi$ $\sigma$
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks evryone I solved it now... I see what I did wrong..

Regards,
\(\displaystyle |\pi\rangle\)