Difference between revisions of "2024 AMC 10B Problems/Problem 6"

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~BenjaminDong01
 
~BenjaminDong01
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==Solution 6 - Overkill==
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Let <math>x</math> denote the length and <math>y</math> denote the width of the rectangle. We have <math>xy=2024</math> and we wish to minimize <math>2x + 2y</math>. WLOG, let <math>x \le y</math>
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From <math>xy = 2024</math>, we have that <math>y = \frac{2024}{x}</math>. So we minimize the expression <math>f(x) = 2x + 2\left(\frac{2024}{x}\right) = 2x + \frac{4048}{x}</math>. The derivative of this expression is <cmath>f'(x) = \frac{d}{dx}\left(2x + \frac{4048}{x}\right) = 2 - \frac{4048}{x^2}.</cmath> We set the derivative equal to zero to find the absolute extrema of <math>f(x)</math>. In other words, <cmath>f'(x) = 2 - \frac{4048}{x^2} = 0.</cmath> Solving this yields <math>x = \pm \sqrt{2024}</math>, which means the perimeter has a maximum when <math>x</math> has this value. Only the positive value of <math>x</math> is feasible since we deal with side lengths. However, we also want <math>x</math> to be an integer.
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We claim that <math>f(x)</math> is a monotonically increasing function when <math>x \le \sqrt{2024}</math>. Looking back at the function of the derivative, <math>f'(x) = 2 - \frac{4048}{x^2}</math>, it is positive when <math>x \le \sqrt{2024}</math>, which means that <math>f(x)</math> is monotonically increasing on this domain.
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Thus, we want <math>x</math> to be the greatest integer less than <math>\sqrt{2024}</math> to maximize the perimeter and satisfy the problem's constraints. So, <math>x = 44</math> and <math>y = \frac{2024}{44} = 46</math>, and the desired perimeter is <math>2(44 + 46) = \boxed{\textbf{(B) }180}</math>.
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please don't do this in the actual competition.
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~rnatog337
  
 
==🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️==
 
==🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️==

Revision as of 16:29, 1 December 2024

Problem

A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?

$\textbf{(A) } 160 \qquad\textbf{(B) } 180 \qquad\textbf{(C) } 222 \qquad\textbf{(D) } 228 \qquad\textbf{(E) } 390$

Solution 1 - Prime Factorization

We can start by assigning the values x and y for both sides. Here is the equation representing the area:


$x \cdot y = 2024$

Let's write out 2024 fully factorized.


$2^3 \cdot 11 \cdot 23$

Since we know that $x^2 > (x+1)(x-1)$, we want the two closest numbers possible. After some quick analysis, those two numbers are $44$ and $46$. $\\44+46=90$

Now we multiply by $2$ and get $\boxed{\textbf{(B) }180}.$

Solution by IshikaSaini.

Solution 2 - Squared Numbers Trick

We know that $x^2 = (x-1)(x+1)+1$ . Recall that $45^2 = 2025$.

If I want 1 less than 2025, which is 2024, I can take 1 number higher and 1 number lower from 45, which are 46 and 44. These are the 2 sides of the minimum perimeter because the 2 numbers are closest to each other, which is what we want to get the minimum.

Finding the perimeter with $2(46+44)$ we get $\boxed{\textbf{(B) }180}.$

Solution by ~Taha Jazaeri

Note: The square of any positive integer with units digit $5$, written in the format $10x + 5$ where $x$ is a positive integer, is equal to $100(x)(x+1)+25$.

~Cattycute

Solution 3 - AM-GM Inequality

Denote the numbers as $x, \frac{2024}{x}$. We know that per AM-GM, $x+\frac{2024}{x} \geq 2\sqrt{2024}$, but since $2\sqrt{2025} = 90$, $2\sqrt{2024}$ must be slightly less than 90, so $2x + 2\frac{2024}{x}$ must be slightly less than 180, eliminating A as a possible answer choice. Proceed with the following solutions above to get 44 and 46, which is $\boxed{\textbf{(B) }180}.$

-aleyang

Solution 4 - Difference of Squares

Note that the year 2025 is a perfect square. The beauty of this year shines down to 2024, which is 1 year lower. $45^2-1^2=2024=(45-1)(45+1)$

Knowing this is the closest possible we can get to a square with only integer factors of 2024, these two are our happy numbers! Add them up to get:

$44+46+44+46=\boxed{\textbf{180}}$ ~BenjaminDong01

Solution 5 - Get Lucky

Note: This is what I did.

$\sqrt{2025}=45$

Assuming it's a square,

$45\cdot4=\boxed{\textbf{180}}$

~BenjaminDong01

Solution 6 - Overkill

Let $x$ denote the length and $y$ denote the width of the rectangle. We have $xy=2024$ and we wish to minimize $2x + 2y$. WLOG, let $x \le y$

From $xy = 2024$, we have that $y = \frac{2024}{x}$. So we minimize the expression $f(x) = 2x + 2\left(\frac{2024}{x}\right) = 2x + \frac{4048}{x}$. The derivative of this expression is \[f'(x) = \frac{d}{dx}\left(2x + \frac{4048}{x}\right) = 2 - \frac{4048}{x^2}.\] We set the derivative equal to zero to find the absolute extrema of $f(x)$. In other words, \[f'(x) = 2 - \frac{4048}{x^2} = 0.\] Solving this yields $x = \pm \sqrt{2024}$, which means the perimeter has a maximum when $x$ has this value. Only the positive value of $x$ is feasible since we deal with side lengths. However, we also want $x$ to be an integer.

We claim that $f(x)$ is a monotonically increasing function when $x \le \sqrt{2024}$. Looking back at the function of the derivative, $f'(x) = 2 - \frac{4048}{x^2}$, it is positive when $x \le \sqrt{2024}$, which means that $f(x)$ is monotonically increasing on this domain.

Thus, we want $x$ to be the greatest integer less than $\sqrt{2024}$ to maximize the perimeter and satisfy the problem's constraints. So, $x = 44$ and $y = \frac{2024}{44} = 46$, and the desired perimeter is $2(44 + 46) = \boxed{\textbf{(B) }180}$.

please don't do this in the actual competition.

~rnatog337

🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️

https://youtu.be/T_QESWAKUUk?si=aCvtmf24mzh2HPCJ

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/QLziG_2e7CY?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

Video Solution by Daily Dose of Math

https://youtu.be/k1bGPUrhYE4

~Thesmartgreekmathdude

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 10 Problems and Solutions

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