2023 AMC 12B Problem 11

Question

What is the maximum area of an isosceles trapezoid that has legs of length

1

and one base twice as long as the other?

\dfrac{5}{4}

\dfrac{8}{7}

\dfrac{5\sqrt{2}}{4}

\dfrac{3}{2}

\dfrac{3\sqrt{3}}{4}

Drop the two heights; each leg spans a horizontal distance of

\frac{\text{long base}-\text{short base}}{2}

With shorter base

a,

the area is

\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}};

maximize its square

Let the bases be

a

and

2a.

Each leg has horizontal offset

\tfrac{a}{2},

so the height is

\sqrt{1-\tfrac{a^2}{4}}

and the area is

A=\tfrac{3a}{2}\sqrt{1-\tfrac{a^2}{4}}.

Then

A^2=\tfrac94\left(a^2-\tfrac{a^4}{4}\right),

maximized when

a^2=2.

There the height is

\tfrac{1}{\sqrt2}

and

A=\tfrac{3\sqrt2}{2}\cdot\tfrac{1}{\sqrt2}=\tfrac{3}{2}.

Thus, the correct answer is D.

Problem 11 in other years