Maximum and Minimum Value on Disk (No Lagrange)

Question

Find the maximum and minimum values of $f(x,y)=5x^2+6y^2$ on the disk $D: x^2+y^2≤1.$

How would I do this question without using Lagrange?

EDIT: How would I do this using x= cos x and y = sinx then inputting them into my equation to solve for theta? EDIT: I get up to $-10(cos(theta)sin(theta) +12(cos(theta)sin(theta)$ = 0

Answer 1

Hint: $5x^2+6y^2=5(x^2+y^2)+y^2$.

Answer 2

$$0\le x^2+y^2 \le 1$$ $$\therefore 0\le 5x^2+5y^2 \le 5$$ $$\therefore 0+y^2 \le 5x^2+6y^2 \le 5+y^2 $$ $$\therefore 0 \le 0+y^2 \le 5x^2 +6y^2 \le 5+y^2 \le 6 $$ $$\therefore 0\le f(x,y) \le 6 $$

Answer 3

The polar parametrization of the disk $x^2+y^2\le1$ is $x=r\cos\theta$ and $y=r\sin\theta$, with $r\le 1$. Then,

$$f(x,y)=g(r,\theta)=r^2(5\cos^2\theta+6\sin^2\theta)=r^2(5+\sin^2\theta)$$

For a given $r$, $g_{\theta}’(r,\theta)=0$ leads to $\sin2\theta=0$. It is straightforward to find that the minimum is
$g(r, 0)=g(r, \pi)=5r^2$ and maximum is $g(r,\frac\pi2)=g(r, \frac{3\pi}2)=6r^2$.

Then, taking $r\in[0,1]$ into account to get the overall minimum $g(0,\theta)=0$ and the overall maximum $g(1,\frac\pi2)=g(1, \frac{3\pi}2)=6$.