Dtyjlui

I have the following two variable function $f(x,y)=2x^2+2y^2+2xy+4x-y$ and I need to find its global extremas on the region $xleq 0, ygeq 0, yleq x+3$.

As in the other questions I asked in the last days, I have no solution to this old exam question so I would like your feedback on wether my results are correct or not.

In the inside of our domain, we have :
$$f_x=4x+2y+4=0$$
$$f_y=4y+2x-1=0$$
So $y=1$ and $x=frac{-3}{2}$ . This point lies in our domain so it is a candidate.

On the border, we have $$f(x, 0)=2x^2+4x, f'=4x+4 = 0 $$ so $x=-1$

We also have $$f(0,y)=2y^2-y, f'=4y-1=0$$ so $y=frac{1}{4}$


And $$f(x, x+3)=2x^2+2(x+3)^2+2x(x+3)+4x-(x+3), f'=3(4x+7)=0$$ (WolframAlpha gets the same derivative for this $f(x,x+3)$). So we get $x=-frac{7}{4}, y=x+3=frac{5}{4}$

Now, we also have our endpoints $(-3,0), (0,3), (0,0)$

Now we plug each of our candidate into our function and look for the greatest/ smallest value which will be our global maximum/ minimum.

$(-frac{3}{2},1) : -frac{7}{2}$
$(-1,0):-2$
$(0,frac{1}{4}) : - frac{1}{8}$
$(- frac{7}{4}, frac{5}{4}) : - frac{27}{8}$
$(-3,0) : 6$
$(0,0) : 0$
$(0,3) : 15$

So there seems to be a global maximum of $15$ at $(0,3)$ and a global minimum of $- frac{7}{2}$ at $(- frac{3}{2},1)$.

Are my results correct or did I miss/ forgot something ?

Also, this problem took me quite some time. Not enormously, but still. Is there a shortcut or a better way to proceed to see that the function actually reaches its global maximum at one of the endpoints and its minimum at $(- frac{3}{2}, 1)$ ?

Thanks for your feedback !

A solution without calculus.

Adopt new variables $x = u - v$, $y = u + v$. The objective function:
begin{align*}
2x^2 + 2y^2 + 2xy + 4x - y &= 2(u^2 - 2uv + v^2) + 2(u^2 + 2uv + v^2) + 2(u^2 - v^2) + 4(u-v) - (u+v) \
&= 6u^2 + 2v^2 + 3u - 5v \
&= 6 left( u + frac{1}{4} right)^2 + 2 left(v - frac{5}{4}right)^2 - frac{7}{2}.
end{align*}
The constraints:
begin{align*}
x leq 0 &Longrightarrow u leq v \
y geq 0 &Longrightarrow u geq -v \
y leq x + 3 &Longrightarrow v leq frac{3}{2}.
end{align*}
We may combine the constraints as an easier-to-visualize $|u| leq v leq frac{3}{2}$.

The objective function clearly has a global minimum over all of $mathbb{R}^2$ at $(u, v) = (-frac{1}{4}, frac{5}{4})$, i.e. $(x, y) = (-frac{3}{2}, 1)$, which is within the constraints. Furthermore, as the objective function is convex, its maximum on a convex polygon must be at a vertex, and it is easy to check all three possible vertices. The maximum is $(u, v) = (frac{3}{2}, frac{3}{2})$ or $(x, y) = (0, 3)$.