2023 AMC 12B Problems/Problem 11
Contents
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
Let the trapezoid be with . Extend and to meet at point . Then, notice with side length ratio and . Thus, and .
The problem reduces to maximizing the area of , an isosceles triangle with legs of length . Analyzing the sine area formula, this is clearly maximized when , so and
-PIDay
Solution 2
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 3 (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 just find the minimum of what's inside the square root (since the square root function is increasing). Take the derivative of , This is equal to zero at but the solution must be positive so .
Solution 4 (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
Solution 5
Denote and as the two bases. We can straightforwardly find that the height of the trapezoid is by the Pythagorean theorem.
The area of this trapezoid is then given by the expression
Let be the maximum value that this expression achieves. Since is positive and assuming , we can perform the following operations to maximize , thus , and thus . We have
The maximum vaue of this quartic occurs at , or when
It follows that the area of the trapezoid is equal to
-Benedict T (countmath1)
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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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