For what $x$ and $y$ polynomial has maximum value?

Question

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: $x=1$, $y=2$

Answer 1

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...

Answer 2

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 $(1,2)$.

Answer 3

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 $16$.

Answer 4

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.

Answer 5

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, $$c=16.$$