Answer
maximum $f(20)= 1250$
Work Step by Step
Step 1. Given $f(x)=-3x^2+120x+50$, $a_2=-3\lt0$, so there is a maximum.
Step 2. The maximum happens at $x=-\frac{120}{2(-3)}=20$ and we have $f(20)=-3(20)^2+120(20)+50=1250$
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