2020 AMC 12B Problems/Problem 25
Contents
Problem 25
For each real number with , let numbers and be chosen independently at random from the intervals and , respectively, and let be the probability that
What is the maximum value of
Solution
Let's start first by manipulating the given inequality.
Let's consider the boundary cases: and
Solving, we get and . Solving the second case gives us and . If we graph these equations in , we get a rhombus shape. Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement.
From the region graph, notice that in order to maximize , . We can solve the rest with geometric probability.
When consists of a triangle with area and a trapezoid with bases and and height . Finally, to calculate , we divide this area by , so
After expanding out, we get . In order to maximize this expression, we must minimize .
By AM-GM, , which we can achieve by setting .
Therefore, the maximum value of is
-Solution by Qqqwerw
Video Solution
On The Spot STEM: https://www.youtube.com/watch?v=5goLUdObBrY
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
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