Đáp án:
$P_{max}=5\sqrt{5}$ khi $(x;y)=\left( \dfrac{9}{5};\dfrac{4}{5}\right)$
Giải thích các bước giải:
Giả thiết: $16x^2-72xy+81y^2+x^2+9y^2=9$
$⇔9=(4x-9y)^2+x^2+9y^2 \geq x^2+9y^2$
Ta có:
$P^4=(3\sqrt{x}+4.\sqrt{4y})^4 \leq \left((3^2+4^2)(x+4y) \right)^2=625.(x+4y)^2$
$⇒P^4 \leq 625\left(1.x+\dfrac{4}{3}·3y\right)^2 \leq 625\left(1^2+\left( \dfrac{4}{3}\right)^2 \right)(x^2+9y^2)$
$⇒P^4 \leq \dfrac{15625}{9}(x^2+9y^2) \leq \dfrac{15625}{9}.9=15625$
$⇒P \leq \sqrt[4]{15625}=5\sqrt{5}$
$P_{max}=5\sqrt{5}$ khi $(x;y)=\left( \dfrac{9}{5};\dfrac{4}{5}\right)$
