Max and min of a polynomial on the unit circle

Question

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

Answer 1

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 $c$.

Answer 2

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

Answer 3

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