Originally Posted by

**MilK** i got these 2 eqn for partial x and y

for partial x = 2pi( -2x(x+y) +a^2 -x^2 -4y^2)

for partial y = 2pi( -8y(x+y) +a^2 -x^2 -4y^2)

subtract these two eqn,

i get 2pi(-6x)(x+y) = 0 Mr F says: This is wrong.

The correct result is $\displaystyle 2\pi [-8y(x + y) - -2x(x + y)] = 0 \Rightarrow (x + y)(2x - 8y) = 0 \Rightarrow (x + y) (x - 4y) = 0$ ..........

Therefore x = -y OR x = 4y. x = 4y is the one you want since x = -y is impossible since x> 0 and y > 0. So sub x = 4y into either of your original equations and solve for y. once you have y, get x. Then test the nature of the solutions.

therefore x+y = 0 and -12pi(x) =0

solving i get x=0, y=0

the question stated : find the max value of this eqn :

2pi(x+y)(a^2 - x^2 -4y^2)

something is wrong somewhere, after finding fxx=0, fyy=0 and fxy=0, (0,0) is not the max point.

what when wrong?