Question

cho a, b là các số thực dương thỏa mãn 2a+3b≤4

Tìm Min M=\(\dfrac{2003}{a}\)+\(\dfrac{2016}{b}\)+3000a−5502b

Answer 1

\(M=2003\left(\dfrac{1}{a}+4a\right)+2016\left(\dfrac{1}{b}+b\right)-5012a-7518b\)

\(M=2003\left(\dfrac{1}{a}+4a\right)+2016\left(\dfrac{1}{b}+b\right)-2506\left(2a+3b\right)\)

\(M\ge2003.2\sqrt{\dfrac{4a}{a}}+2016.2\sqrt{\dfrac{b}{b}}-2506.4=2020\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{1}{2};1\right)\)