max of x+y, of an quarter annulus ( (r1)^2 =< x^2+y^2 =< (r2)^2 where r2>r1

[tender acorn]

confused on bit where x has to = y too

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[tender acorn]

x and y are not negative

[grand moth]

what you can potentially do is use polar coordinates you can set $x=rcos(t)$ and $y=rsin(t)$ for $r \in [r_1,r_2]$ and $t \in [0, \frac{\pi}{2}]$ and study for which values of $r$ and $t$ is $f(r,t)=rcos(t)+rsin(t)$ maximum

[jovial tiger]

Rotor 😑

[tender acorn]

erm this question is done without th eneed of all this but ok

i got the answer anyways

the maximum is on y=x

but

I have another question

for a quarter annulus between and inclusive of 1 and 4 whats the maximum of x^2+y^2-4xy I found the maximum of xy at the lowest border and the minimum of x^2+y^2 which is -1, to get 1-4(1/2) but this is wrong as the answer is -4

[inland lance]

Well, you can do it as usual: find the critical points in the interior, then on the smooth parts of the boundary. Then, compare the values at those critical points and at the non-smooth parts of the boundary.

[tender acorn]

I got it nvm ty tho i just had to try both cases

[autumn sandal]

@tender acorn has given 1 rep to @inland lance

[tender acorn]

+close