Difference between revisions of "2023 AMC 12B Problems/Problem 11"

Revision as of 23:24, 15 November 2023

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?

$\textbf{(A) }\frac 54 \qquad \textbf{(B) } \frac 87 \qquad \textbf{(C)} \frac{5\sqrt2}4 \qquad \textbf{(D) } \frac 32 \qquad \textbf{(E) } \frac{3\sqrt3}4$

Solution 1

Denote by $x$ the length of the shorter base. Thus, the height of the trapezoid is $\begin{align*} \sqrt{1^2 - \left( \frac{x}{2} \right)^2} . \end{align*}$

Thus, the area of the trapezoid is $\begin{align*} \frac{1}{2} \left( x + 2 x \right) \sqrt{1^2 - \left( \frac{x}{2} \right)^2} & = \frac{3}{4} \sqrt{x^2 \left( 4 - x^2 \right)} \\ & \leq \frac{3}{4} \frac{x^2 + \left( 4 - x^2 \right)}{2} \\ & = \boxed{\textbf{(D) } \frac{3}{2}} , \end{align*}$

where the inequality follows from the AM-GM inequality and it is binding if and only if $x^2 = 4 - x^2$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2 (Calculus)

Derive the expression for area $A = \frac{3}{4}x\sqrt{4-x^2}$ as in the solution above. To find the minimum, we can take the derivative with respect to $x$ : $\frac{dA}{dx} = \frac{3}{4}\sqrt{4-x^2}+\left(\frac{3x}{4}\right)\frac{-2x}{2\sqrt{4-x^2}} = \frac{6-3x^2}{2\sqrt{4-x^2}}.$ This expression is equal to zero when $x=\pm\sqrt{2}$ , so $A$ has two critical points at $\pm\sqrt{2}$ . But given the bounds of the problem, we can conclude $x = \sqrt{2}$ maximizes $A$ (alternatively you can do first derivative test). Plugging that value back in, we get $A_{\text{max}} = \boxed{(\text{D})\ \frac{3}{2}}$ .

~cantalon

(Slightly Simpler)

Or rewrite the expression for area to be

$A = \frac{3}{4}\sqrt{4x^2-x^4}$

Now to find the minimum, we can take the derivative of what is inside the square root. $f'(x)=8x-4x^3$ This is equal to zero at $x=\pm\sqrt{2}$ but the solution must be positive so $x=\sqrt{2}$ . Putting that into the formula for area, we got $A = \boxed{(\text{D})\ \frac{3}{2}}$ .

Solution 3 (Trigonometry)

Let the length of the shorter base of the trapezoid be $2x$ and the height of the trapezoid be $y$ .

$[asy] unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.5, 0.5), D=(-0.5, 0.5); draw(A--B--C--D--cycle, black); label("$2x$",(0,0.58),(0,0)); label("$2x$",(0,-0.08),(0,0)); label("$x$",(-0.75,-0.08),(0,0)); label("$x$",(0.75,-0.08),(0,0)); draw(D--(-0.5,0),black); draw(C--(0.5,0),black); label("$y$",(0.58,0.25)); label("$y$",(-0.42,0.25)); [/asy]$

Each leg has length $1$ if and only if $x^2+y^2=1$ , where $x$ and $y$ are positive real numbers. The general solution to this equation is $(x,y)=(\cos t,\sin t)$ for any number $0<t<\frac{\pi}{2}$ so that $x$ and $y$ are positive. The area to maximize is $\frac{1}{2}(2x+4x)y=3xy$ Hence, we maximize $3\sin t\cos t$ for $0<t<\frac{\pi}{2}$ . $\begin{align*} 3xy &= 3\sin t\cos t \\ &= \frac{3}{2}(2\sin t\cos t) \\ &= \frac{3}{2}\sin(2t) \end{align*}$ The maximum of $\sin(2t)$ is $1$ , thus the maximum of $3xy$ is $\boxed{\frac{3}{2}}$ which occurs at $t=\frac{\pi}{4}$ , satisfying the inequality $0<t<\frac{\pi}{2}$ .

~Robabob1

@@ Line 47: / Line 47: @@
 <cmath>f'(x)=8x-4x^3</cmath>
 This is equal to zero at <math>x=\pm\sqrt{2}</math> but the solution must be positive so <math>x=\sqrt{2}</math>. Putting that into the formula for area, we got <math>A = \boxed{(\text{D})\ \frac{3}{2}}</math>.
+==Solution 3 (Trigonometry)==
+Let the length of the shorter base of the trapezoid be <math>2x</math> and the height of the trapezoid be <math>y</math>.
+<asy>
+unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.5, 0.5), D=(-0.5, 0.5);  draw(A--B--C--D--cycle, black);  label("$2x$",(0,0.58),(0,0)); label("$2x$",(0,-0.08),(0,0)); label("$x$",(-0.75,-0.08),(0,0)); label("$x$",(0.75,-0.08),(0,0)); draw(D--(-0.5,0),black); draw(C--(0.5,0),black); label("$y$",(0.58,0.25)); label("$y$",(-0.42,0.25));
+</asy>
+Each leg has length <math>1</math> if and only if <math>x^2+y^2=1</math>, where <math>x</math> and <math>y</math> are positive real numbers. The general solution to this equation is <cmath>(x,y)=(\cos t,\sin t)</cmath> for any number <math>0<t<\frac{\pi}{2}</math> so that <math>x</math> and <math>y</math> are positive. The area to maximize is <cmath>\frac{1}{2}(2x+4x)y=3xy</cmath> Hence, we maximize <math>3\sin t\cos t</math> for <math>0<t<\frac{\pi}{2}</math>.
+<cmath>
+\begin{align*}
+xy &= 3\sin t\cos t \\
+    &= \frac{3}{2}(2\sin t\cos t) \\
+    &= \frac{3}{2}\sin(2t)
+\end{align*}
+</cmath>
+The maximum of <math>\sin(2t)</math> is <math>1</math>, thus the maximum of <math>3xy</math> is <math>\boxed{\frac{3}{2}}</math> which occurs at <math>t=\frac{\pi}{4}</math>, satisfying the inequality <math>0<t<\frac{\pi}{2}</math>.
+~Robabob1
 ==See Also==
 {{AMC12 box|year=2023|ab=B|num-b=10|num-a=12}}
 {{MAA Notice}}

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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