For what $x$ and $y$ polynomial has maximum value? 
For what $x,y\in\mathbb R$ does the polynomial
  $$-5x^2-2xy-2y^2+14x+10y-1$$
  attain a maximum?

My attempt:
I called $\alpha$ maximum value.
$$-5x^2-2xy-2y^2+14x+10y-1\leqslant\alpha$$
$$-5x^2-2xy-2y^2+14x+10y-1-\alpha\leqslant 0$$
$$5x^2+2xy+2y^2-14x-10y+1+\alpha\geqslant 0$$
$$(x+y)^2+(y-5)^2+3x^2+(x-7)^2-73+\alpha\geqslant0$$
$$\alpha\geqslant73$$
So the lowest maximum value turned out to be $73$, but after checking answers I was wrong-maximum is $16$, so my further plans to calculate from that $x$ and $y$ seemed purposless. I'd like to see solution using only high school knowledge.
Ans: $x=1$, $y=2$
 A: Write $$f(x)=-5x^2-(2y-14)x-2y^2+10y-1$$ This quadratic function on $x$ with parameter $y$ achieves maximum at 
$$p=-{b\over 2a} = {2y-14\over -10}$$ and this maximum is $$ q= -{b^2-4ac\over 4a} = {(2y-14)^2+20(-2y^2+10y-1)\over 20}=$$
So you need to find the maximum of $$g(y)= (2y-14)^2+20(-2y^2+10y-1)$$
$$ =-36y^2+144y +176$$
And now you can repeat the procedure which was done for $x$.
The maximum is at $y= -{144\over -72}=2$ (and thus $x=1$) and that maximum value is...
A: Your idea is quite correct, but you have to complete squares in a different way:
\begin{gather}
-(5x^2+2xy+2y^2-14x-10y+1) =\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +2y^2-10y+1-\frac{1}{5}y^2-\frac{49}{5}+\frac{14}{5}y\right]=\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +\frac{9}{5}y^2-\frac{36}{5}y-\frac{44}{5}\right]=\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +\left(\frac{3}{\sqrt 5}y-\frac{6}{\sqrt 5}\right)^2-\frac{36}{5}-\frac{44}{5}\right]=\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +\left(\frac{3}{\sqrt 5}y-\frac{6}{\sqrt 5}\right)^2-16\right]=\\
-\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 -\left(\frac{3}{\sqrt 5}y-\frac{6}{\sqrt 5}\right)^2+16 \leq 16
\end{gather}
In the first passage I complete the squares in order to eliminate the terms in $x^2$ and $xy$. Then complete in order to eliminate the term in $y^2$ and $y$. So we showed that our function is less or equal to $16$. Imposing the two squares equal to zero you obtain the maximum $16$ and if you do the computation you obtain the point $(1,2)$ that makes it.
This is a standard method to do this kind of exercise. You have to complete squares in a similar way as I did; then you obtain a inequality and putting the squares equal to zero you will find the minimum and the point that makes it.
A: If you know a bit of multivariate calculus you can compute the maximum by taking partial derivatives:
$$\nabla=(-10x-2y+14,-4y-2x+10)=(0,0)$$
Solving, we get $x=1,y=2$ as the only extremum point, where the value is $16$. To prove this is indeed a maximum, note that the partial derivatives are decreasing in the neighbourhood of $(1,2)$.
A: Let the maximum occur at $(a,b)$. If we translate the equation by this amount
$$-5x^2-2xy-2y^2+14x+10y-1$$
becomes
$$-5(u+a)^2-2(u+a)(v+b)-2(v+b)^2+14(u+a)+10(v+b)-1.$$
The linear terms in the expansion are
$$(-10a-2b+14)u+(-2a-4b+10)v.$$
If we cancel the two coefficients, which is achieved by $a=1,b=2$, the function reduces to
$$-5u^2-2uv-2v^2+c$$ where $c$ is some constant term.
Now, as the discriminant of the quadratic terms is negative, 
$$-5u^2-2uv-2v^2\le0$$ and equality only occurs at the origin $u=v=0$.

By expanding the constant term, $$c=16.$$

A: let $k$ be a maximal value.
Thus, for any $x$ and $y$ we have $$-5x^2-2xy-2y^2+14x+10y-1\leq k$$ or
$$5x^2+2(y-7)x+2y^2-10y+1+k\geq0,$$ which gives
$$(y-7)^2-5(2y^2-10y+1+k)\leq0$$ or
$$9y^2-36y-44+5k\geq0,$$ which gives $$18^2-9(-44+5k)\leq0,$$ which gives $$k\geq16.$$
The equality occurs for $$y=\frac{36}{2\cdot9}=2$$ and $$x=-\frac{2(2-7)}{2\cdot5}=1,$$ which says that a minimal value of $k$, 
for which the first inequality is true for any values of $x$ and $y$, it's $16$.
