maximum and minimum value of the function $ f(x,y) = x^{2} + y^{2} + xy + 3(x+y) $ at $ A = \left \{ (x,y): x^{2}+y^{2} \leq 1, x+y\geq 0 \right \} $

Question

Find the maximum and minimum value of the function $ f(x,y) = x^{2} + y^{2} + xy + 3(x+y) $ at $ A = \left \{ (x,y): x^{2}+y^{2} \leq 1, x+y\geq 0 \right \} $ and the points at which it is obtained.

Since the region $x^{2}+y^{2} \leq 1$ is closed, $f$ has a maximum and a minimum, which is either at the boundary or at the critical points of the function.

• $ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0 \Leftrightarrow \left\{\begin{matrix} 2x+y+3=0\\ 2y+x+3=0 \end{matrix}\right. \Leftrightarrow ... \Leftrightarrow \left\{\begin{matrix} y=-1\\ x=-1 \end{matrix}\right. $ (invalid answer)
• Lagrange theorem

$g(x,y)=x^{2}+y^{2}=1$

$x,y \neq 0 , \bigtriangledown f(x,y) = \lambda \bigtriangledown g(x,y) $ and from this $ 3 \times 3 $ system I got $ (x-y)(2-1-2\lambda )=0 $ so $x=y$ and the only valid answer is $ (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) $ (I think) or $ \lambda =\frac{1}{2} $ and the system has no real solutions. That's my try. I don't know if I am right or not. I only found one point. How can I find if this point gives maximum or minimum value?

Answer 1

Solving the system gives $x=y=-1$, not $x=-5,y=1$, but that is an invalid answer as well...

It's clear that switching $x \leftrightarrow y$ gives the same problem back, so the answers will be symmetric in $x$ and $y$.

Indeed, if the extrema are not inside, they must be on the boundary. Your argument covers the case when you are on the circle, but there is also a possibility you are on the line $y=-x$, in which case you have $$ f(x,-x) = x^2, $$ which reaches minimum at $x=0$ (yielding the point $(0,0)$ and maximum at both extremes, $(1/\sqrt{2}, -1/\sqrt{2})$ and $(-1/\sqrt{2}, 1/\sqrt{2})$.

So we have four points, to test them, we can just plug them into $f$ and calculate...

Answer 2

$$x^2+y^2+xy+3(x+y)=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+3(x+y)\geq0.$$ The equality occurs for $x=y=0,$ which saus that we got a minimal value.

Now, by AM-GM and C-S we obtain: $$x^2+y^2+xy+3(x+y)\leq x^2+y^2+\frac{x^2+y^2}{2}+3\sqrt{(1^2+1^2)(x^2+y^2)}\leq1.5+3\sqrt2.$$ The equality occurs for $x=y=\frac{1}{\sqrt2},$ which says that we got a maximal value.