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Let $a,b>0$,and such $$a^2+b^2-ab=4$$ Find the range $$f(a,b)=2a+b-3ab$$

I try let $a=x+y,b=x-y$,then $$a^2+b^2-ab=4\Longrightarrow x^2+3y^2=16,x>y,x>-y$$ so we Let

$$\begin{align} x &=4\cos{t}, \qquad y=\dfrac{4}{\sqrt{3}}\sin{t} \\ f(a,b) &= 2a+b-3ab \\ &= \dfrac{3x}{2}+\dfrac{y}{2}-\frac{3}{4}x^2+\dfrac{3}{4}y^2\\ &=6\cos{t}-\dfrac{2}{\sqrt{3}}\sin{t}-12\cos^2{t}+4\sin^2{t} \end{align}$$

Then I stuck

Hosein Rahnama
  • 14,882
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    Which mathematical means are allowed? Calculus in general? Lagrange multipliers? Just analytic geometry (since your equation defines an ellipse)? And what work have you done on the problem so far? Where are you stuck? – Rory Daulton Dec 19 '15 at 14:37
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    Thanks for showing your work. It seems that you want a solution not using calculus, since your techniques are those of analytic geometry. Is that correct? – Rory Daulton Dec 19 '15 at 14:44
  • One option : http://mathworld.wolfram.com/LagrangeMultiplier.html – lab bhattacharjee Dec 19 '15 at 14:53

1 Answers1

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I think you have made a mistake in your computations as I did the arithmetic with my CAS. However, if we make the substitutions

$$\begin{align} a &= 2 \cos t + \frac{2}{\sqrt{3}} \sin t = \frac{4}{\sqrt{3}} \sin(t+\frac{\pi}{3}) \\ b &= 2 \cos t - \frac{2}{\sqrt{3}} \sin t = \frac{4}{\sqrt{3}} \sin(t+\frac{2\pi}{3}) \end{align}$$

then the constraint equation

$$a^2+b^2-ab=4$$

will be satisfied identically $4=4$. Also, to satisfy $a \gt 0$ and $b \gt 0$ we require that

$$\begin{cases} 0 \lt t+\frac{\pi}{3} \lt \pi \\ 0 \lt t+\frac{2\pi}{3} \lt \pi \end{cases} \to \begin{cases} -\frac{\pi}{3} \lt t \lt \frac{2\pi}{3} \\ -\frac{2\pi}{3} \lt t \lt \frac{\pi}{3} \end{cases} \to -\frac{\pi}{3} \lt t \lt \frac{\pi}{3}$$

So, it remains to work on $f(a,b)=2a+b- 3ab$. Now, if we do the substitution in $f(a,b)$ we will get

$$\begin{align}g(t) &= f(a(t),b(t)) \\ &=\frac{2}{\sqrt{3}}\sin t + 6 \cos t - 16 \cos^2 t +4 \\ \end{align}$$

Finally, we can use any method in calculus to find the Maximum and Minimum of this continuous function on the interval $(-\frac{\pi}{3},\frac{\pi}{3})$. So the range of $f(a,b)$ with the desired constraints on $a$ and $b$ will be obtained. In fact, we just reduced a constrained multi-variable optimization problem to a usual single variable optimization problem in calculus.

Hosein Rahnama
  • 14,882
  • The final calculations seem to be tedious. Minimum, maximum – A.Γ. Dec 19 '15 at 16:04
  • @A.G.: Computations in Optimization problems are always tedious to me! :D – Hosein Rahnama Dec 19 '15 at 16:07
  • I mean you are right, but the single variable optimization does not seem to have a nice analytical solution, which looks suspicious. Either something is wrong with the original problem (typo etc) or there is an easier way to solve it. – A.Γ. Dec 19 '15 at 16:16
  • @A.G.: What do you mean by suspicious? It should have a Max and Min in that interval! :) However, there may still be some elegant ways to solve the problem easier. :) – Hosein Rahnama Dec 19 '15 at 16:18
  • "Suspicious" in the sense that it looks like a high school problem that normally is supposed to have a short and beautiful answer (well, at least it was during those happy days when I went to high school myself :-)) – A.Γ. Dec 19 '15 at 16:25
  • @A.G.: Maybe you are right! :) However, high-school students do not know about conic section transformations and it seems that the OP knows them as He\She makes such a change of variables! :) – Hosein Rahnama Dec 19 '15 at 16:29