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

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~cantalon
 
~cantalon
 +
 +
(Slightly Simpler)
 +
 +
Or rewrite the expression for area to be
 +
 +
<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.
 +
<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>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2023|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2023|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:35, 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 $A = \boxed{(\text{D})\ \frac{3}{2}}$.

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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