Divide the number 84 into two parts such that the product of one part and square of the other is maximum.Solution : Let one part be x then the other part will be \(84-x\).\(\therefore f(x)=\square\)\(\therefore f'(x)=168x-3x^2\)For extreme values \(f…

Q.3 (C) (i) Divide the number 84 into two parts such that the product of one part and square of the other is maximum.

Solution : Let one part be x then the other part will be \(84-x\).

\(\therefore f(x)=\square\)
\(\therefore f'(x)=168x-3x^2\)
For extreme values \(f'(x)=0\)
\(168x-3x^2=0\)
\(\therefore 3x(56-x)=0\)
\(\therefore x=\square\) OR \(\square\)
\(f''(x)=168-6x\)
If \(x=0,\ f''(0)=168-6(0)=168\gt0\)
\(\therefore\) function attains minimum at \(x=0\)
If \(x=56,\ f''(56)=\square\lt0\)
\(\therefore\) function attains maximum at \(x=56\)
\(\therefore\) Two parts of 84 are \(\square\) and \(\square\)