If $x$ and $y$ are two real quantities connected by the equation $9x^2+2xy+y^2-92x-20y+244=0$ then will $x \in [3,6]$, and $y \in [1,10]$?

Question

If $x$ and $y$ are two real quantities connected by the equation $$9x^2+2xy+y^2-92x-20y+244=0$$ then will $x \in [3,6]$ , and $y \in [1,10]$ ?
What could be some of the intuitive / standard ways of finding that ?

What I've tried:

My attempt using completing squares :

The given $eq^n$ can be expressed as $8(x-3)(x-6) + (x+y-10)^2 = 0$

From here , it's clear that $(x+y-10)^2 ≥ 0$ and for that $(x-3)(x-6) ≤ 0$ which implies $x \in [3,6]$ and $|x+y-10| ≥ 0$

Now since, $|x+y-10| ≥ 0$ and $x \in [3,6]$ therefore, $y \in [4,7]$ . Where am I being wrong ?

Answer 1

The given $eq^n$ can be expressed as $8(x-3)(x-6) + (x+y-10)^2 = 0$

From here , it's clear that $(x+y-10)^2 ≥ 0$ and for that $(x-3)(x-6) ≤ 0$ which implies $x \in [3,6]$

That works. Or, along the same idea, just with a different turn at the end:

$$ \begin{align} 0 &= 9x^2 + 2xy + y^2 - 92x - 20y + 244 \\ &= y^2 + 2(x - 10)y \color{red}{+ (x - 10)^2 - (x - 10)^2} + 9x^2 - 92x + 244 \\ &= (y + x - 10)^2 + 8 x^2 - 72 x + 144 \\ &= (x + y - 10)^2 + 8 \left(x - \frac{9}{2}\right)^2 - 18 \\ \iff &\quad(x + y - 10)^2 + 8 \left(x - \frac{9}{2}\right)^2 = 18 \end{align} $$

It follows that:

$$ 8 \left(x - \frac{9}{2}\right)^2 \le 18 \;\iff\; \left|x - \frac{9}{2}\right| \le \sqrt{\frac{18}{8}} = \frac{3}{2} \;\iff\; 3 = \frac{9}{2} - \frac{3}{2}\le x \le \frac{9}{2} + \frac{3}{2} = 6 $$

[ EDIT ] At this point, knowing the range of $x$, it may be tempting to try and use the same argument for $|x + y - 10| \le \sqrt{18} = 3\sqrt{2}$ $\iff 10 - 3\sqrt{2} - x \le y \le 10 + 3\sqrt{2} - x$ $\implies y \in [4 - 3\sqrt{2}, 7 + 3\sqrt{2}]$. This is not wrong, but the bounds are not tight, either, and the interval is not the true range of $y$ because some values can never be attained. That's because the two terms $(x + y - 10)$ and $x - \dfrac{9}{2}$ are not independent, and the extrema of $y$ are not attained at the same points where the extrema of $x$ are attained. In fact, as shown below, the actual range of $y$ is $[1,10] \subsetneqq [4 - 3\sqrt{2}, 7 + 3\sqrt{2}]$.

since, $|x+y-10| ≥ 0$ and $x \in [3,6]$ therefore, $y \in [4,7]$. Where am I being wrong?

The "therefore" step is not justified, and the implication is wrong. The inequality $|x+y-10| \ge 0$ holds true no matter what the values of $x,y$ are, because an absolute value is always non-negative.

But, remember the idea that worked the first time around: you completed the square in $y$, and you found the range of $x$. Now, try to complete the other square in $x$, and you'll get the range of $\,y\,$:

$$ \begin{align} 0 &= 9x^2 + 2xy + y^2 - 92x - 20y + 244 \\ &= (3x)^2 + 2\,\frac{y-46}{3} \, 3x \color{red}{+ \left(\frac{y-46}{3}\right)^2 - \left(\frac{y-46}{3}\right)^2} + y^2 - 20y + 244 \\ &= \left(3x + \frac{y-46}{3}\right)^2 + \frac{8}{9} y^2 - \frac{88}{9}y + \frac{80}{9} \\ &= \left(3x + \frac{y-46}{3}\right)^2 + \frac{8}{9} (y-1)(y-10) \end{align} $$

It follows that:

$$ (y-1)(y-10) \le 0 \;\;\iff\;\; 1 \le y \le 10 $$

Answer 2

If one is allowed to use the Lagrange multiplier then this is fairly standard.

Let $F(x,y,\lambda) = x + \lambda(9x^2+2xy+y^2-92x-20y+244)$, then $$F'_x = 2\lambda(9x+y-46)+1, F'_y = 2\lambda(x+y-10),$$ so letting $F'_x = F'_y = 0$ we get $x = (72\lambda-1)/(16\lambda), y = (88\lambda+1)/(16\lambda)$, and $0 = 9x^2+2xy+y^2-92x-20y+244 = (1 - 576\lambda^2)/(32\lambda^2)$, which implies that $\lambda = \pm 1/24$, and this corresponds to $x=3$ or $x=6$.

Similarly, let $G(x,y,\lambda) = y + \lambda(9x^2+2xy+y^2-92x-20y+244)$, then $$G'_x = 2\lambda(9x+y-46), G'_y = 2\lambda(x+y-10)+1,$$ so $G'_x = G'_y = 0$ yields $x = (72\lambda+1)/(16\lambda), y = (88\lambda-9)/(16\lambda)$, and $0 = 9x^2+2xy+y^2-92x-20y+244 = (9 - 576\lambda^2)/(32\lambda^2)$, so $\lambda = \pm 1/8$, and this corresponds to $y=1$ or $y=10$.

Answer 3

Trigonometric substitution should help (as Jianing Song commented).

We can write $$9x^2+2xy+y^2-92x-20y+244=0$$ as $$y^2+2(x-10)y+9x^2-92x+244=0$$ $$(y+x-10)^2-(x-10)^2+9x^2-92x+244=0$$ $$(y+x-10)^2+8x^2-72x+144=0$$ $$(x + y - 10)^2 + 8 \left(x - \frac{9}{2}\right)^2 = 18$$

Now, dividing the both sides by $18$, we get $$\frac{1}{18}(x + y - 10)^2 + \frac 49 \left(x - \frac{9}{2}\right)^2 = 1$$ which can be written as $$\bigg(\frac{x+y-10}{3\sqrt 2}\bigg)^2+\bigg(\frac 23x-3\bigg)^2=1$$

Here, let us use trigonometric substitution : $$\cos t=\frac{x+y-10}{3\sqrt 2},\qquad \sin t=\frac 23x-3\tag1$$ Solving $(1)$ for $x,y$ gives $$x=\frac 32\sin t+\frac 92$$ and $$y=3\sqrt 2\ \cos t-\frac 32\sin t+\frac{11}2=\frac 92\cos(t+\alpha)+\frac{11}{2}$$ where $\alpha$ is such that $\cos\alpha=\frac{2\sqrt 2}{3}$ and $\sin\alpha=\frac 13$.

Therefore, we finally obtain $$3=\frac 32(-1)+\frac 92\leqslant x\leqslant \frac 32\cdot 1+\frac 92=6$$ and $$1=\frac 92(-1)+\frac{11}{2}\leqslant y\leqslant \frac 92\cdot 1+\frac{11}{2}=10$$

Answer 4

GEOMETRIC APPROACH

Quadratic equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ represents a conic section, which may be degenerate. The given equation is $9x^2+2xy+y^2-92x-20y+244=0.$

With the use of matrix representation, compute the determinant of the associate matrix $$\left|{\begin{matrix}A & B/2 & D/2\\B/2 & C & E/2\\ D/2 & E/2 & F\end{matrix}}\right|=\left|{\begin{matrix}9 & 1 & -46\\1 & 1 & -10\\ -46 & -10 & 244\end{matrix}}\right|=-144\neq0,$$ and $$\left|{\begin{matrix}A & B/2 \\B/2 & C\end{matrix}}\right|=\left|{\begin{matrix}9 & 1 \\1 & 1 \end{matrix}}\right|=8>0,$$

therefore $9x^2+2xy+y^2-92x-20y+244=0$ is equation of an ellipse. It is enclosed in a rectangle with horizontal and vertical sides, which lie on horizontal and vertical tangents to this ellipse, respectively.

Horizontal tangents
We can use a simple high-school approach: find a horizontal line with equation $y=c$ that has a unique common point with the ellipse. This leads to the quadratic equation $$9x^2+x(2c-92)+(c^2-20c+244)=0,$$ whose discriminant $$\Delta=-16(2c^2-22c+20)$$ must be $0.$
The two horizontal tangents have equations $$y=1,\quad y=10.$$ It follows necessarily that $$1\le y\le10$$ for each point on the ellipse, because ellipse is convex.

A similar computation gives two vertical tangents $$x=3, \quad x=6,$$ which proves that $$3\le x\le 6$$ for any point on the ellipse.

Answer 5

Let $$F(x.y)=9x^2+2xy+y^2-92x-20y+244=0,$$ then $$\dfrac{\text dy}{\text dx}=-\dfrac{F'_x}{F'_y}= \dfrac{18x+2y-92}{2x+2y-20}.$$ $$\dfrac{\text dx}{\text dy}=-\dfrac{F'_y}{F'_x}= \dfrac{2x+2y-20}{18x+2y-92}.$$ I.e., extremes by $y$ are defined by the system $$\begin{cases} 18x+2y-92=0\\ 9x^2+2xy+y^2-92x-20y+244=0, \end{cases}$$ $$\begin{cases} y=46-9x\\ 9x^2+2x(46-9x)+y(46-9x)^2-92x-20(46-9x)+244=0, \end{cases}$$ $$\begin{cases} y=46-9x\\ 72(x-4)(x-5)=0, \end{cases}$$ $$\dbinom xy\in\left\{\dbinom4{10},\dbinom51\right\},$$ $$\color{brown}{\mathbf{y\in[1,10].}}$$

Extremes by x are defined by the system $$\begin{cases} 2x+2y-20=0\\ 9x^2+2xy+y^2-92x-20y+244=0, \end{cases}$$ $$\begin{cases} x=10-y\\ 9(10-y)^2+2y(10-y)+y^2-92(10-y)-20y+244=0 \end{cases}$$ $$\begin{cases} x=10-y\\ 8(y-4)(y-7)=0, \end{cases}$$ $$\dbinom xy\in\left\{\dbinom64,\dbinom37\right\},$$ $$\color{brown}{\mathbf{x\in[3,6].}}$$

Easily to see that the center of the given figure is the point $\,\left(\dfrac92, \dfrac{11}2\right).\,$ This allows to present it in the form of $$F(x,y)=9\left(x-\dfrac92\right)^2+2\left(x-\dfrac92\right)\left(y-\dfrac{11}2\right)+\left(y-\dfrac{11}2\right)^2-18=0.$$ or, going to main axes, $$\dfrac{2\left(\left(x-\dfrac92\right) -\big(\sqrt{17}+4\big)\left(y-\dfrac{11}2\right)\right)^2}{9(153+37\sqrt{17})} +\dfrac{2\left(\big(\sqrt{17}+4\big)\left(x-\dfrac92\right) +\left(y-\dfrac{11}2\right)\right)^2}{9(17+3\sqrt{17})}=1,$$ $$\dfrac{\left(\left(x-\dfrac92\right) -\big(\sqrt{17}+4\big)\left(y-\dfrac{11}2\right)\right)^2}{a^2} +\dfrac{\left(\big(\sqrt{17}+4\big)\left(x-\dfrac92\right) +\left(y-\dfrac{11}2\right)\right)^2}{b^2}=1.$$ where $$a=\sqrt{\frac92(153+37\sqrt{17})}\approx 37.080953,$$ $$b=\sqrt{\frac92(17+3\sqrt{17})}\approx 11.496170.$$ I.e., the given figure is a sloped ellipse.

Answer 6

First, we have $$0 = 9x^2+2xy+y^2-92x-20y+244 = 8(x-3)(x-6) + (x + y - 10)^2$$ which results in $8(x - 3)(x - 6) \le 0$. Thus, $x \in [3, 6]$.

Second, we have $$0 = 9x^2+2xy+y^2-92x-20y+244 = 9 (x - 5)^2 + (1 - y)^2 + (1 - y)(18 - 2x)$$ which results in $1 - y \le 0$ (using $18 - 2x > 0$). Thus, $y \ge 1$.

Third, we have $$0 = 9x^2+2xy+y^2-92x-20y+244 = 9(x - 4)^2 + (y - 10)^2 + 2x (y - 10)$$ which results in $y - 10 \le 0$ (using $2x > 0$). Thus, $y \le 10$.

Thus, $x\in [3, 6]$ and $y \in [1, 10]$.

We are done.

Answer 7

From what you found, namely $8(x-3)(x-6) + (x+y-10)^2 = 0$, we have $$y=10-x\pm\sqrt{8(3-x)(x-6)}.$$ Then, $y'=-1\pm\frac{36-8x}{\sqrt{72x-8x^2-144}}=0$ gives $x=4$ and $x=5$ as interior extrema and which in turn give the values $y=2,10$ and $y=9,1$ respectively. Hence, $y\in[1,10].$