Max and Min value of a function on a circle

Question

Find the maximum and minimum values of the function $f(x,y) = 5x^2 + 2xy + 5y^2$ on the circle $x^2 + y^2 = 1$.

After substituting the equation of the circle in that of the function and then equating $f'(x) = 0$, I get the values of $y$ to be $\pm1\sqrt2$. Plugging these into the function, the resulting values are $\max(f) = 6$ and $\min(f) = 4$. However, the answer given is $\max(f) = \min(f) = 5$. I would like to know where I went wrong.

Answer 1

The given answer is wrong. For example $f(x,y) \leq 5x^{2}+(x^{2}+y^{2})+5y^{2} \leq 6$ and the value $6$ is attained when $x=y=\frac 1 {\sqrt 2}$. Hence the correct maximum value is $6$. Note also that minimum and maximum cannot be the same for a non-constant function.

Answer 2

You can do the following:

Let $x= cos\theta$ and $y= sin\theta.$ Then the function $f$ becomes $f(\theta)$= 5 + $sin(2\theta)$. Then we're done. What's the max and min of sine function? ( Recall that $sin2\theta =2(sin\theta)(cos\theta)$).

Max f = 6 and min f = 4

Answer 3

On circle $f(x,y) = 2xy+5 = (x+y)^2+4$ and product $xy$ on circle have maximum in $x=y=\frac{1}{\sqrt{2}}$ and minimum $x=1,y=-1$ or $x=-1,y=1$. So 6 and 4 are maximum and minimum.

Answer 4

$$x^2+y^2\pm2xy\ge0\iff \pm2xy\ge-(x^2+y^2)\ge1$$

If $xy>0; 2xy\ge1$

If $xy<0;-2xy\ge1\iff2xy\le-1$