Here, use the below function:
\(f\left( x \right) = - 3{x^2} + 120x + 50\)
First find the critical value as follow:
\(\begin{array}{c}f'\left( x \right) = 0\\f'\left( x \right) = - 6x + 120\\0 = - 6x + 120\end{array}\)
\(\begin{array}{c}6x - 120 = 0\\6\left( {x - {\rm{20}}} \right) = 0\\\left( {x - {\rm{20}}} \right) = 0\\x = 20\end{array}\)
\(f\left( x \right) = - 3{x^2} + 120x + 50\)
First find the critical value as follow:
\(\begin{array}{c}f'\left( x \right) = 0\\f'\left( x \right) = - 6x + 120\\0 = - 6x + 120\end{array}\)
\(\begin{array}{c}6x - 120 = 0\\6\left( {x - {\rm{20}}} \right) = 0\\\left( {x - {\rm{20}}} \right) = 0\\x = 20\end{array}\)
So, the maximum or minimum value is,
\(\begin{array}{c}f\left( {20} \right) = - 3{\left( {20} \right)^2} + 120\left( {20} \right) + 50\\ = - 1200 + 2400 + 50\\ = 1250\end{array}\)