2020 AMC 12B Problem 22

Split the fraction: $\dfrac{(2^t - 3t)t}{4^t} = \dfrac{t}{2^t} - \dfrac{3t^2}{4^t}.$ Let $u = \dfrac{t}{2^t},$ so $\dfrac{t^2}{4^t} = u^2$ and the expression is $u - 3u^2.$ This parabola has maximum at $u = \tfrac16,$ with value $\tfrac16 - 3\cdot\tfrac1{36} = \tfrac1{12}.$ Since $u = \tfrac{t}{2^t}$ is continuous and attains the value $\tfrac16,$ the maximum is achieved. Thus, the correct answer is C.