Question

cho 2a^2+5b^2=11

Tìm max và min N = 3a-2b

Answer 1

\(N^2=\left(3a-2b\right)^2=\left(\dfrac{3}{\sqrt{2}}.\sqrt{2}a-\dfrac{2}{\sqrt{5}}.\sqrt{5}b\right)^2\le\left(\dfrac{9}{2}+\dfrac{4}{5}\right)\left(2a^2+5b^2\right)=\dfrac{583}{10}\)

\(\Rightarrow-\sqrt{\dfrac{583}{10}}\le N\le\sqrt{\dfrac{583}{10}}\)

\(N_{max}=\sqrt{\dfrac{583}{10}}\) 

\(N_{min}=-\sqrt{\dfrac{583}{10}}\)