smc

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