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2023 AMC 12B Problems/Problem 11: Difference between revisions

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<cmath>A = \frac{3}{4}\sqrt{4x^2-x^4}</cmath>
<cmath>A = \frac{3}{4}\sqrt{4x^2-x^4}</cmath>


Now to find the minimum, we can take the derivative of what is inside the square root.
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 <math>4x^2-x^4</math>,
<cmath>f'(x)=8x-4x^3</cmath>
<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>.
This is equal to zero at <math>x=0,\pm\sqrt{2}</math> but the solution must be positive so <math>x=\sqrt{2}</math>.


==Solution 3 (Trigonometry)==
==Solution 3 (Trigonometry)==

Revision as of 12:40, 16 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 just find the minimum of what's inside the square root (since the square root function is increasing). Take the derivative of $4x^2-x^4$, \[f'(x)=8x-4x^3.\] This is equal to zero at $x=0,\pm\sqrt{2}$ but the solution must be positive so $x=\sqrt{2}$.

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 $\boxed{\text{(D) }\frac{3}{2}}$ which occurs at $t=\frac{\pi}{4}$, satisfying the inequality $0<t<\frac{\pi}{2}$.

~Robabob1

Video Solution 1 by OmegaLearn

https://youtu.be/WlXBbaHl-z4


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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