Find the max/min value of $x^2+y^2$ subject to $3x^2+5xy+3y^2=1$

Question

Find the minimum and maximum value of $x^2+y^2$ subject to $3x^2+5xy+3y^2=1$.

That is a precalculus problem. So we are not allowed to use Lagrange multipliers. Here is my approach.

$$3x^2+5xy+(3y^2-1)=0$$

Using quadratic formula

$$x=\frac {-5y\pm \sqrt {25y^2-12(3y^2-1)}}{6}$$

So

$$x=\frac {-5y\pm\sqrt {12-11y^2}}{6}$$

I reduced the number of variables now we have an univariate function.

$$x^2+y^2=y^2+\frac {\big(5y\mp\sqrt {12-11y^2}\big)^2}{36}$$

At least I know that $12\ge 11y^2$ or $y^2\le\frac {12}{11}$. How can I continue from here?

Answer 1

Another way .

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Substituting $\thinspace x=a+b,\thinspace y=a-b\thinspace$, then you have :

$$ \begin{align}&3\left(x^2+y^2\right)+5xy=1\\ \iff &11a^2+b^2=1\end{align} $$

and

$$ \begin{align}M=x^2+y^2=2\left(a^2+b^2\right)\end{align} $$

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This leads to the following :

$\color{#0a0}{{\underline{\color{#c00}{\rm {Global \thinspace \thinspace minimum }}}}}$

$$ \begin{align}&11\left(\frac M2-b^2\right)+b^2=1\\ \iff &M=\frac {2\left(10b^2+1\right)}{11}≥\frac {2}{11}\end{align} $$

$\color{#0a0}{{\underline{\color{#c00}{\rm {Global \thinspace \thinspace maximum }}}}}$

$$ \begin{align}&11a^2+\frac M2-a^2=1\\ \iff &M=2\left(1-10a^2\right)≤2 \end{align} $$

which completes the answer .

Answer 2

The problem may be simpler in polar coordinates. Our constraint is:

$$3x^2+5xy+3y^2=1$$ $$3(x^2+y^2)+5xy=1$$ $$3r^2+5(r \cos \theta)(r \sin \theta)=1$$ $$3r^2+5r^2\cos\theta\sin\theta=1$$ $$3r^2+\frac{5}{2}r^2\sin(2\theta)=1$$ $$r^2 = \frac{2}{6+5\sin(2\theta)}$$

And we want to minimize $x^2 + y^2 = r^2$, which we now have an explicit formula for. Since $\sin(2\theta)$ can take on any value between -1 and 1, the denominator is between 1 and 11. Therefore,

$$\boxed{\frac{2}{11} \le x^2 + y^2 \le 2}$$

Answer 3

$$\frac{11}{4}(x+y)^2+\frac{1}{4}(x-y)^2=1$$

Let $$x+y=\frac{2}{\sqrt{11}}\sin\theta,~~~~x-y=2\cos\theta$$

we get

$$f=x^2+y^2=\frac{(x+y)^2+(x-y)^2}2=\frac{2}{11}\sin^2\theta+2\cos^2\theta$$ hence $$f=x^2+y^2=\frac{2}{11}+\frac{20}{11}\cos^2\theta$$ so we get $$\min f=\frac{2}{11},~~~\max f=2$$

Answer 4

By homogeneity, these min and max are the respective reciprocals of the maximum and minimum values of the Rayleigh quotient $$\begin{align}\frac{3x^2+5xy+3y^2}{x^2+y^2}&=3\cos^2t+5\cos t\sin t+3\sin^2t\\&=3+\frac52\sin(2t),\end{align}$$ so $$\min=\frac1{3+\frac52}=\frac2{11},\quad\max=\frac1{3-\frac52}=2.$$

Answer 5

Min: $3x^2+5xy+3y^2=1 \Leftrightarrow 3(x^2+y^2)=1-5xy \Leftrightarrow 3(x^2+y^2)-1=-5xy$(*)

We have: $\frac{x^2+y^2}{2}\ge xy(\text{Because we can mutiply both side by 2 and then we have: } x^2+y^2\ge2xy\to x^2-2xy+y^2=(x-y)^2\ge 0,\text{equality hold when } x=y(1)$

Apply (1) for $5xy:$ $$\to5xy\le\frac{5(x^2+y^2)}{2} \\\Leftrightarrow-5xy\ge-\frac{5(x^2+y^2)}{2} \\\Leftrightarrow1-5xy\ge1-\frac{5(x^2+y^2)}{2}$$ from (*):$$\Rightarrow1+3x^2+3y^2-1\ge\frac{2-5x^2-5y^2}{2} \\\frac{2+6x^2+6y^2-2}{2}\ge\frac{2-5x^2-5y^2}{2}\\ \Leftrightarrow2+6x^2+6y^2-2\ge2-5x^2-5y^2 \\\Leftrightarrow11(x^2+y^2)\ge2\\ \Leftrightarrow x^2+y^2\ge\frac{2}{11}$$

The equality hold when $x^2+y^2=\frac{2}{11}$ , $x=y$ and $3x^2+5xy+3y^2=1\to x=y=\pm\frac{1}{\sqrt{11}}$

Here is my approach, but I haven't figured out for the max value.

Answer 6

Find the $\min/\max$ value of $x^2+y^2$ when $3x^2+5xy+3y^2=1$.

The shape of $3x^2+5xy+3y^2=1$ will be an eclipse below:

So, we can draw a circle with the center at $(0, 0)$ which passes the marked points, like:

And it's well-known that the marked point is the meets of the eclipse and $y=\pm x$.

So, the radiuses of the two circles each are $\dfrac{1}{\sqrt{11}}$ and $1$. ($6x^2\pm5x^2=1.$)

Therefore, $2\left( \dfrac{1}{\sqrt{11}} \right)^2 \leq x^2+y^2 \leq 2\cdot 1^2$.

$\min(x^2+y^2) = \dfrac{2}{11}, \max(x^2+y^2) = 2.$

Answer 7

Just another way using AM-GM, $x^2+y^2\geqslant 2|xy| \implies \frac52(x^2+y^2)\geqslant 5xy \geqslant -\frac52(x^2+y^2)$.

$$\implies (3+\tfrac52)(x^2+y^2)\geqslant3(x^2+y^2)+5xy=1 \geqslant (3-\tfrac52) (x^2+y^2)$$ $$\implies \frac1{3-\frac52}\geqslant x^2+y^2\geqslant \frac1{3+\frac52} \iff 2\geqslant x^2+y^2\geqslant \frac2{11}$$ Equality is possible when $x=-y=\pm 1$ for the upper bound, and when $x=y=\pm\frac1{\sqrt{11}}$ for the lower bound, so these bounds are in fact the max and min respectively.

Answer 8

There are already 7 diffrent solutions. I provide the 8th one at precalculus level (even with no trigonometry, AM-GM,...).

Denote $k = x^2 +y^2$. If $y = 0$ then $k = x^2 +y^2 = x^2 = \frac{1}{3}$ (later, we observe in $(2)$ that it is not minimum or maximum)

Suppose $y \ne 0$, denote $t = \frac{x}{y}$, we have

$$ k= \frac{x^2+y^2}{3x^2+5xy+3y^2}=\frac{t^2+1}{3t^2+5t+3} \Longrightarrow (3k-1)t^2 + 5kt+(3k-1) = 0 \tag{1} $$

$(1)$ has solution if and only if $$\begin{align} \Delta =(5k)^2-4(3k-1)^2 \ge 0 \iff (k-2)(11k-2) \le 0 \iff \color{red}{\frac{2}{11} \le k \le 2} \tag{2} \end{align}$$ The minimum $k = \frac{2}{11}$ is reached if and only if $t = -\frac{5k}{2(3k-1)} = 1$ or $x = y = \pm \frac{1}{\sqrt{11}} $

The maximum $k = 2$ is reached if and only if $t = -\frac{5k}{2(3k-1)} = -1$ or $x = -y = \pm 1 $