MHB Finding Max/Min Values for f(x,y)=x^2+y^2+2x-2y+2

[Petrus]

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

 

[HallsofIvy]

Decide max and min value for
$$f(x,y)=x^2+y^2+2x-2y+2$$ where $$x^2+y^2 \leq1$$
progress:
for solving that equation $$f_x(x,y), f_y(x,y)$$ we get the point $$(1,-1)$$

But [math](-1)^2+ 1^2= 2> 1[/math] so (1, -1) is not in the desired region and is irrelevant.

if we do parametric $$x= \cos(t) \, y= \sin(y)$$ where $$0\leq t \leq 2\pi$$
if we put those value and simplify and derivate we get
$$-2\sin(t)-2\cos(t)=0 <=> 1=-\tan(t)$$
that means its on second and 4th quadrants

"On second and 4th quadrants" includes a lot of territory! $$tan(t)= -1$$ when $$t= 3\pi/4$$ and $$T= -\pi/4$$

do I take $$ -tan^{-1}$$ now?
and get now got $$-tan^{-1}(t)$$
so we got $$x=\cos(-\tan^{1}(1)) <=> t=2$$,

It's hard to tell what you mean. "$$-tan^{-1}$$" is meaningless without an argument. I take it you mean $$tan^{-1}(-1)$$ which gives the values I give above: $$t= 3\pi/4$$ and $$t= -\pi/4$$

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

Regards,
$$|\pi\rangle$$

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

 

[chisigma]

Decide max and min value for
$$f(x,y)=x^2+y^2+2x-2y+2$$ where $$x^2+y^2 \leq1$$
progress:
for solving that equation $$f_x(x,y), f_y(x,y)$$ we get the point $$(1,-1)$$

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]

Thanks evryone I solved it now... I see what I did wrong..

Regards,
$$|\pi\rangle$$