Difference between revisions of "2023 AMC 12B Problems/Problem 11"
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Revision as of 02:10, 16 November 2023
Contents
[hide]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 .
Solution 3 (Trigonometry)
Let the length of the shorter base of the trapezoid be and the height of the trapezoid be .
Each leg has length if and only if , where and are positive real numbers. The general solution to this equation is for any number so that and are positive. The area to maximize is Hence, we maximize for . The maximum of is , thus the maximum of is which occurs at , satisfying the inequality .
~Robabob1
Video Solution 1 by OmegaLearn
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.