I'm required to comment on the maximum and minimum values of the function :
$f(x,y)=2x^2+2xy+y^2-2x+2y+2$
My question is, if both $x$ and $y$ are variables, how can I use standard methods to find the range on this function? Is there a trick here?
Also, how can we use differentiation in it(if we can)?
On
The maximum does not exist.
Let $k$ is a minimal value.
Thus, $2x^2+2(y-1)x+y^2+2y+2-k\geq0$, which is quadratic inequality of $x$
with positive coefficient before $x^2$.
Thus, we need $\Delta\leq0$ or $$(y-1)^2-2y^2-4y-4+2k\leq0$$ or $$y^2+6y+3-2k\geq0,$$ which is quadratic inequality again.
Thus, we need $$3^2-(3-2k)\leq0$$ or $k\leq-3$ and from here $k=-3$ it's a maximal value of $k$ for which the inequality $$2x^2+2xy+y^2-2x+2y+2\geq k$$ is true for all reals $x$ and $y$.
For $k=-3$ we obtain the equality case: $x=-\frac{2(y-1)}{2\cdot2}$ and $y=-3$.
Thus, $-3$ is a minimal value.
Done!
Since we know the answer already, we can write this solution by the following "nice" way.
For $x=2$ and $y=-3$ we get a value $-3$.
Thus, it's enough to prove that $$2x^2+2xy+y^2-2x+2y+2\geq-3$$ or $$(2x+y-1)^2+(y+3)^2\geq0,$$ which is obvious.
Done!
HINT: it is$$2x^2+2xy+y^2-2x+2y+2\geq -3$$ and the equal sign holds for $x=2,y=-3$ let $$f(x,y)=2x^2+2xy+y^2-2x+2y+2$$ and compute the partial derivatives and solve the System $$4x+2y-2=0$$ and $$2x+2y+2=0$$