If-x-2-y-2-9-4a-2-9b-2-16-then-maximum-value-of-4a-2-x-2-9b-2-y-2-12abxy-is-

Question Number 32629 by rahul 19 last updated on 29/Mar/18

I f x^{2} + y^{2} = 9, 4 a^{2} + 9 b^{2} = 16,

t h e n m a x i m u m v a l u e o f

4 a^{2} x^{2} + 9 b^{2} y^{2} - 12 a b x y i s ?

Answered by MJS last updated on 31/Mar/18

x^{2} + y^{2} = 9 \Rightarrow ∣ x ∣⩽ 3 \land ∣ y ∣⩽ 3

4 a^{2} + 9 b^{2} = 16 \Rightarrow ∣ a ∣⩽ 2 \land ∣ b ∣⩽ \frac{4}{3}

4 a^{2} x^{2} + 9 b^{2} y^{2} - 12 a b x y = {(2 a x - 3 b y)}^{2}

y = \pm \sqrt{9 - x^{2}}

b = \pm \frac{2}{3} \sqrt{4 - a^{2}}

{(2 a x \pm 2 \sqrt{4 - a^{2}} \sqrt{9 - x^{2}})}^{2} =

= 4 {(a x \pm \sqrt{4 - a^{2}} \sqrt{9 - x^{2}})}^{2}

f (x) = a x + \sqrt{4 - a^{2}} \sqrt{9 - x^{2}}

f^{'} (x) = a - \frac{x \sqrt{4 - a^{2}}}{\sqrt{9 - x^{2}}}

f^{'} (x) = 0 \Rightarrow x = \pm \frac{3}{2} a

4 {(a x \pm \sqrt{4 - a^{2}} \sqrt{9 - x^{2}})}^{2} =

= 4 {(\pm \frac{3}{2} a^{2} \pm \frac{3}{2} (4 - a^{2}))}^{2} =

= 4 {(\pm 6)}^{2} \lor 4 {(\pm (6 - 3 a^{2}))}^{2} =

= 144 \lor 36 {(2 - a^{2})}^{2}

g (a) = 36 {(2 - a^{2})}^{2}

g^{'} (a) = 144 a (2 - a^{2})

g^{'} (a) = 0 \Rightarrow a = 0 \lor a = \pm \sqrt{2}

g (\pm \sqrt{2}) = 0

g (0) = 144

Answer = 144

a = 0

b = \pm \frac{4}{3}

x = 0

y = \pm 3

Commented by rahul 19 last updated on 31/Mar/18

t h a n k u s i r!