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#### 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,

\(\displaystyle |\pi\rangle\)

\(\displaystyle f(x,y)=x^2+y^2+2x-2y+2\) 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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