Question

cho x>0 tim Max \(A=\dfrac{1200x\left(12+x\right)}{\left(48+16x\right)^2}\)

Answer 1

\(A=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\);\(A>0\forall x>0\)

Gọi \(A_0\in MGT\) của A

\(\Rightarrow A_0=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\) có nghiệm

\(\Rightarrow A_0\left(12+4x\right)^2=75x\left(12+x\right)\)

\(\Leftrightarrow x^2\left(16A_0-75\right)+x\left(96A_0-900\right)+144A_0=0\) có nghiệm

\(\Leftrightarrow\Delta\ge0\Leftrightarrow-4A_0+25\ge0\)\(\Leftrightarrow A_0\le\dfrac{25}{4}\)

\(\Rightarrow maxA=\dfrac{25}{4}\)