Problem:
What is the maximum value of 4t(2t−3t)t for real values of t ?
Answer Choices:
A. 161
B. 151
C. 121
D. 101
E. 91 Solution:
We proceed by using AM−GM. We get 2(2t−3t)+3t≥(2t−3t)(3t). Thus, squaring gives us that 4t−1≥(2t−3t)(3t). Remembering what we want to find, we divide both sides of the inequality by the positive amount of 3⋅4t1. We get the maximal values as (C)121, and we are done.
OR
Set u=t2−t. Then the expression in the problem can be written as
R=−3t24−t+t2−t=−3u2+u=−3(u−1/6)2+121≤121
It is easy to see that u=61 is attained for some value of t between t=0 and t=1, thus the maximal value of R is (C)121.
Solution 3 (Calculus Needed) We want to maximize f(t)=4t(2t−3t)t=4tt⋅2t−3t2. We can use the first derivative test. Use quotient rule to get the following: