What is the maximum value of ((2^t - 3t) t)/(4^t) for real values of t?

2020 AMC 12B Problem 22 — Solution

2020 AMC 12B Problem 22

Below is the professionally curated solution for Problem 22 of the 2020 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

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Concepts:substitution quadratic optimization

Difficulty rating: 1860

22.

What is the maximum value of

$\frac{(2^t - 3t)\,t}{4^t}$

for real values of $t?$

$\dfrac{1}{16}$

$\dfrac{1}{15}$

$\dfrac{1}{12}$

$\dfrac{1}{10}$

$\dfrac19$

Solution:

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.

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