Difference between revisions of "2023 AMC 12B Problems/Problem 11"
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<cmath>A = \frac{3}{4}x\sqrt{4-x^2}</cmath> | <cmath>A = \frac{3}{4}x\sqrt{4-x^2}</cmath> | ||
as in the solution above. To find the minimum, we can take the derivative with respect to <math>x</math>: | as in the solution above. To find the minimum, we can take the derivative with respect to <math>x</math>: | ||
− | <cmath>\frac{dA}{dx} = \frac{3}{4}\sqrt{4-x^2} | + | <cmath>\frac{dA}{dx} = \frac{3}{4}\sqrt{4-x^2}+\frac{3x}{4}\frac{-2x}{2\sqrt{4-x^2}} = \frac{6-3x^2}{2\sqrt{4-x^2}}.</cmath> |
This expression is equal to zero when <math>x=\pm\sqrt{2}</math>, so <math>A</math> has two critical points at <math>\pm\sqrt{2}</math>. But given the bounds of the problem, we can conclude <math>x = \sqrt{2}</math> maximizes <math>A</math> (alternatively you can do first derivative test). Plugging that value back in, we get <math>A_{\text{max}} = \boxed{(\text{D})\ \frac{3}{2}}</math>. | This expression is equal to zero when <math>x=\pm\sqrt{2}</math>, so <math>A</math> has two critical points at <math>\pm\sqrt{2}</math>. But given the bounds of the problem, we can conclude <math>x = \sqrt{2}</math> maximizes <math>A</math> (alternatively you can do first derivative test). Plugging that value back in, we get <math>A_{\text{max}} = \boxed{(\text{D})\ \frac{3}{2}}</math>. | ||
Revision as of 20:26, 15 November 2023
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
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.