Lagrange multipliers to find min and max

[nazmi]

hi there.. trying to solved this problem.. i solved it until i got max=45 and min = 5
however.. my tutor said my answer was wrong and ask me to check it by myself..
i cant figure out where did i do wrong..

The temperature at a point (x, y) on a metal plate is T(x, y) = 4x^2 ? 4xy + y^2 .
An ant, walking on the plate, traverses a circle of radius 5 centered at the origin.
Using the method of Lagrange multipliers, find the highest and lowest
temperatures encountered by the ant.

i've done until
when y = 2x,
i substitute into x^2+y^2=25
i got x=+-(5)^1/2

when x=-2y,
i substitute into x^2+y^2=25
i got y= +-(5)^1/2

so.. i got my critical points..
---> [(5^1/2) , (5^1/2)]
---> [(5^1/2) , -(5^1/2)]
---> [-(5^1/2) , (5^1/2)]
---> [-(5^1/2) , -(5^1/2)]
my critical points are correct? :?:

 

[galactus]

$\displaystyle x^{2}+y^{2}=25$ is the constraint; $\displaystyle 8x-4y=2x{\lambda}, \;\ -4x+2y=2y{\lambda}; \;\ \frac{4x-2y}{x}=\frac{-2x+y}{y}, \;\ 2x^{2}+3xy-2y^{2}=0, \;\ (2x-y)(x+2y)=0$, $\displaystyle y=2x \;\ or \;\ x=-2y$

If $\displaystyle y=2x, \;\ then \;\ x^{2}+(2x)^{2}=25, \;\ x=\pm\sqrt{5}$

If $\displaystyle x=-2y, \;\ then \;\ (-2y^{2})+y^{2}=25, \;\ y=\pm\sqrt{5}.$

$\displaystyle T(-\sqrt{5}, \;\ -2\sqrt{5})=T(\sqrt{5}, \;\ 2\sqrt{5})=0$

and $\displaystyle T(2\sqrt{5}, \;\ -\sqrt{5})=T(-2\sqrt{5},\sqrt{5})=125$

The highest temperature is 125 and the lowest is 0.