1962 AHSME Problems/Problem 26
Problem
For any real value of the maximum value of
is:
Solution
Let Since
is a quadratic and the quadratic term is negative, the maximum will be
when written in the form
. We see that
, and so
. Plugging in this value yields $f(\dfrac{4}{3}) = \dfrac{32}{3}-3 \cdot \dfrac{16}{9} = \dfrac{32}{3} - \dfrac{16}{3} = \boxed{$ (Error compiling LaTeX. Unknown error_msg)\text{E}\ \dfrac{16}{3}}$