11
$\begingroup$

I have been trying to solve this problem,

For $x, y \in \mathbb{R}$ such that $x^2+y^2=1$, find the minimum and maximum value of $$(3x+2y)^2+(x+2y)^2$$

There are many ways to solve this problem such as

  • Trigonometry (substitute $x = \sin\theta, y = \cos\theta, \theta\in\left[0, 2\pi\right)$)

  • Lagrange Multipliers

  • Solve for $y$ and plug it in to the wanted expression and differentiate it.

  • If we let the wanted expression equal to some constant $k$, it is an ellipse thus use linear transformation to rotate and find values of $k$ such that the ellipse is circumscribed inside/outside the circle

Using trigonometry, I found out that the answer is min: $9-\sqrt{65}$, max: $9+\sqrt{65}$.

But this problem actually came up in an algebra textbook (no relation to calculus or even pre-calculus), with the topic of Quadratic Equations and Discriminants.

Considering the context of the textbook, I am trying to find a solution using discriminants. My first try was to let the given expression $(3x+2y)^2+(x+2y)^2 = k$ and assume that the min/max will happen when the discriminant is greater than or equal to $0$. (Since the equation must have real roots).

So what I did was expand to get $10x^2+16xy+8y^2=k$, use $x^2+y^2=1$ to simplify into $$2x^2+16xy+8-k=0$$ this equation of $x$ must have real roots so $D/4\geq0$. $$64y^2-2(8-k)\geq0$$

Now I am currently stuck here. Is there a way to get around this? I would really like to find a solution that uses discriminants. Thanks in advance.


Note: I just mentioned the $4$ other ways to solve this problem so that I can emphasize that I would really like to see a solution using discriminants.

Also, any solution without calculus (or with only topics covered until high school - maybe -) are always welcome.

$\endgroup$
  • 1
    $\begingroup$ Nice question. +1 Thanks for the good formatting! $\endgroup$ – Adrian Keister Feb 26 '18 at 15:15
7
$\begingroup$

Let $$f(x,y)=(3x+2y)^2+(x+2y)^2=10x^2+16xy+8y^2.$$ If $f(x,y)=a$ for some $x$, $y$ on the unit circle then $$f(x,y)-a(x^2+y^2)=0\tag1$$ has nonzero real solutions. But if $(1)$ has a nonzero real solution, one can scale it so that $x^2+y^2=1$, and then $f(x,y)=a$ for that $(x,y)$.

But $$f(x,y)-a(x^2+y^2) =(10-a)x^2+16xy+(8-a)y^2.$$ This has a nonzero real solution iff the discriminant $$16^2-4(10-a)(8-a)\ge0.$$ Solve this inequality for $a$ to get the values of $f$ on the unit circle.

$\endgroup$
  • $\begingroup$ Can I suggest a small clarifying edit: just after equation (1), change "has nonzero real solutions" to "for any fixed $(x, y)$, has a nonzero real solution for $a$." $\endgroup$ – John Hughes Feb 26 '18 at 15:30
  • 2
    $\begingroup$ Wow.. I am baffled.. I never thought I could change $a$ to $a(x^2+y^2)$. Thank you very much, you just solved my problem that troubled me for a week! I can finally go to sleep now! Thanks! $\endgroup$ – zxcvber Feb 26 '18 at 15:31
  • $\begingroup$ Very nice! Reminiscent of Lagrange. $\endgroup$ – Adrian Keister Feb 26 '18 at 15:37
3
$\begingroup$

$$y=10\sin^2t+16\sin t\cos t+8\cos^2t$$

Divide both sides by $\cos^2t$ and set $\tan t=a$

$$(1+a^2)y=10a^2+16a+8\iff a^2(10-y)+16a+8-y=0$$ which is a Quadratic Equation in $y$

So, the discriminant must be $\ge0$

$$\implies16^2\ge4(10-y)(8-y)\iff(y-9)^2\le64+1\iff-\sqrt{65}\le y-9\le\sqrt{65}$$

$\endgroup$
  • $\begingroup$ I like the way you changed $\tan t = a$. $\endgroup$ – zxcvber Feb 26 '18 at 16:25
1
$\begingroup$

Substituting with trig functions gives, $$y=10\sin^2t + 16\sin t\cos t+8\cos^2t$$ Using the identity $\sin^2x+\cos^2x=1$, it simplifies to $$y=8 + 2\sin^2t + 16\sin t \cos t$$ Use the double-angle formula $\sin{2x}=2\sin x \cos x$ and half-angle formula $\sin^2t = \frac{1-\cos{2t}}{2}$. $$y=9+8\sin 2t-\cos 2t$$

Now, $$y = 9+\sqrt{65}\left(\frac{8}{\sqrt{65}} \sin 2t - \frac{1}{\sqrt{65}} \cos 2t\right)$$ $$y=9+\sqrt{65} \sin{(2t-\alpha)}$$ where $\cos\alpha=8/\sqrt{65}, \sin\alpha=1/\sqrt{65}$

Thus maximum is $9+\sqrt{65}$, minimum is $9-\sqrt{65}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.