Problem:
What is the maximum area of an isosceles trapezoid that has legs of length 1 and one base twice as long as the other?
Answer Choices:
A. 45
B. 78
C. 452
D. 23
E. 433
Solution:
Let the bases have lengths x and 2x. Extend the legs of the isosceles trapezoid to form two isosceles triangles with vertex angle θ, as shown. with vertex angle θ, as shown.
Because the ratio of the bases is 1:2, the two congruent sides of the smaller isosceles triangle have length 1 . The area of the trapezoid is 43 the area of the large isosceles triangle, so its area is maximized when the area of the large isosceles triangle is maximized. The area of the large isosceles triangle as a function of θ is 21⋅22⋅sinθ=2sinθ. The maximum area of this isosceles triangle is 2 , occurring when θ=2π. Therefore the maximum area of the trapezoid is 43⋅2=(D)23.
OR
Let h be the height of the trapezoid, let x and 2x be the base lengths, and let α be the base angle, as shown.
Then sinα=h and 2cosα=x. The area of the trapezoid is 2x+2x⋅h, which can be written as a function of α :
3cosα⋅sinα=23sin(2α).
Because sin(2α)≤1, the maximum area is (D)23, which occurs when sin(2α)=1 and α=4π.
OR
Let h be the height of the trapezoid, and let x and 2x be the base lengths (see the figure for the second solution). Then 12=h2+(2x)2, so
h=1−(2x)2
The area A of the trapezoid is\
so
A=2x+2x1−(2x)2
A2=49x2(1−4x2)=49x2−169x4.
Let u=x2. Then A2=−169u2+49u, which has a maximum when
.jpg)
