2023 AMC 12B Problems/Problem 11
Problem
What is the maximum area of an isosceles trapezoid that has legs of length and one base twice as long as the other?
Solution 1
Denote by the length of the shorter base. Thus, the height of the trapezoid is
Thus, the area of the trapezoid is
where the inequality follows from the AM-GM inequality and it is binding if and only if .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Calculus)
Derive the expression for area as in the solution above. To find the minimum, we can take the derivative with respect to : This expression is equal to zero when , so has two critical points at . But given the bounds of the problem, we can conclude maximizes (alternatively you can do first derivative test). Plugging that value back in, we get .
~cantalon
(Slightly Simpler)
Or rewrite the expression for area to be
Now to find the minimum, we can take the derivative of what is inside the square root. This is equal to zero at but the solution must be positive so . Putting that into the formula for area, we got .
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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