Algebra
Q.1
$m^2 + n^2 = 5$, find the max value of $3m + 2n$.
Solution 1
Let $x = 3m + 2n, y = 2m - 3n$, we have $x^2 + y^2 = 13m^2+13n^2 = 65$.
So $x^2 = 65 - y^2 = 65 - (2m -3n)^2$.
When $2m = 3n$, $x$ has max value $\sqrt{65}$.
Solution 2
Let $f(m) = 3m + 2n = 3m + 2\sqrt{5 - m^2}$.
So $f'(m) = 3 - \frac{2m}{\sqrt{5 - m^2}}$.
Let $f'(m) = 0$, we get $m^2 = 45/13, n^2=20/13$.
So $max(3m + 2n) = 3\sqrt{45/13} + 2\sqrt{20/13} = \sqrt{65}$
