1962 AHSME Problems/Problem 26

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For any real value of $x$ the maximum value of $8x - 3x^2$ is:

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ \frac{8}3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \frac{16}{3}$


Let $f(x) = 8x-3x^2$ Since $f(x)$ is a quadratic and the quadratic term is negative, the maximum will be $f\bigg(- \dfrac{b}{2a}\bigg)$ when written in the form $ax^2+bx+c$. We see that $a=-3$, and so $- \dfrac{b}{2a} = -\bigg( \dfrac{8}{-6}\bigg) = \dfrac{4}{3}$. Plugging in this value yields $f(\dfrac{4}{3}) = \dfrac{32}{3}-3 \cdot \dfrac{16}{9} = \dfrac{32}{3} - \dfrac{16}{3} = \boxed{\text{E}\ \dfrac{16}{3}}$