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

2020 AMC 12B Problem 22

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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

Let

u = \dfrac{t}{2^t};

then

\dfrac{t^2}{4^t} = u^2

The expression becomes

u - 3u^2,

a downward parabola in

u

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},

\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.