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Find the maximum and minimum of $ f(x,y)= 3x^2+5xy-4y^2$ on the unit circle.

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closed as off-topic by Jared, O.L., Rick Decker, Daniel Rust, mrf Jul 24 '13 at 0:52

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3 Answers

Hint: $x = \cos(\theta)$, $y = \sin(\theta)$. Use addition formulas for $\sin$ and $\cos$ to write in terms of $\cos(2\theta+c)$ for some $c$.

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Have you tried substituting $x = \cos \theta$ and $y = \sin \theta$, and then setting the derivative to $0$?

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Without using calculus and formula of multiple angles we can find the extreme values of $Ax^2+Bxy+Cy^2$ on the unit circle using simple algebra.

Putting $x=\cos t,y=\sin t$ we get,

$$Ax^2+Bxy+Cy^2=A\cos^2t+B\sin t\cos t+C\sin^2t$$

Let $$A\cos^2t+B\sin t\cos t+C\sin^2t=P$$

Dividing either sides by $\cos^2t, A+B\tan t+C\tan^2t=P\sec^2t$

or, $A+B\tan t+C\tan^2t=P(1+\tan^2t)$

$$\implies (C-P)\tan^2t+B\tan t+A-P=0$$ which is a quadratic equation in $\tan t$

As, $t$ is real so is is $\tan t,$ hence the discriminant $$B^2-4(A-P)(C-P)\ge 0$$

Now, $$B^2-4(A-P)(C-P)$$ $$=B^2-4AC-\{4P^2-4P(A+C)\}$$ $$=B^2-4AC+(A+C)^2-\{4P^2-4P(A+C)+(A+C)^2\}$$ $$=B^2+(A-C)^2-\{2P-(A+C)\}^2$$

$$\implies B^2+(A-C)^2-\{2P-(A+C)\}^2\ge0$$

$$\implies \{2P-(A+C)\}^2\le B^2+(A-C)^2$$

$$\implies -\sqrt{B^2+(A-C)^2}\le 2P-(A+C)\le \sqrt{B^2+(A-C)^2} $$ as $y^2\le a^2\implies -a\le y\le a$

$$\implies \frac{A+C-\sqrt{B^2+(A-C)^2}}2\le P\le \frac{A+C+\sqrt{B^2+(A-C)^2}}2$$

Here $A=3,B=5,C=-4$

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^^^ can you explain those steps please? –  user58315 Jan 22 '13 at 17:28
    
@user58315, please point out your doubts? –  lab bhattacharjee Jan 22 '13 at 17:30
    
After dividing both sides by cos^2t I'm lost –  user58315 Jan 22 '13 at 17:37
    
Maybe I'm missing the trig calculation you made. Can you explain how you got p=sec^2t=p(1+tan^2t)? –  user58315 Jan 22 '13 at 17:43
    
@user58315, I think you meant $P\sec^2t=P(1+\tan^2t)$ which is true as $\frac P{\cos^2t}=P\sec^2t$ and $1+\tan^2t=\sec^2t$ –  lab bhattacharjee Jan 22 '13 at 17:57
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