Question

Let x and y be real numbers satisfying 9x2 +16y2 = 1. Then (x+y) is maximum when

Answer 1

$9 {x}^{2} + 16 {y}^{2} = 1$

$\frac{ {x}^{2} }{ \frac{1}{9} } + \frac{ {y}^{2} }{ \frac{1}{16} } = 1$

it is an ellipse with

$a = \frac{1}{3} \: \: and \: \: \: b = \frac{1}{4}$

let any point on it be

$(a cos \alpha \: \: \: \: \: bsin \alpha ) = ( \frac{1}{3} cos \alpha \: \: \: \: \frac{1}{4} sin \alpha )$

$x = \frac{1}{3} cos \alpha \: \: \: and \: \: \: \: \: \: \: y = \frac{1}{4} sin \alpha$

$x + y = \frac{cos \alpha }{3} + \frac{sin \alpha }{4}$

$max(x + y) = \sqrt{ { \frac{1}{3} }^{2} + { \frac{1}{4} }^{2} }$

$\sqrt{ \frac{1}{9} + \frac{1}{16} } = \sqrt{ \frac{16 + 9}{144} } = \sqrt{ \frac{25}{144} } = \frac{5}{12}$