Difference between revisions of "2024 AMC 10B Problems/Problem 6"
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~BenjaminDong01 | ~BenjaminDong01 | ||
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+ | ==Solution 6 - Overkill== | ||
+ | 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. | ||
+ | |||
+ | ~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
Contents
[hide]- 1 Problem
- 2 Solution 1 - Prime Factorization
- 3 Solution 2 - Squared Numbers Trick
- 4 Solution 3 - AM-GM Inequality
- 5 Solution 4 - Difference of Squares
- 6 Solution 5 - Get Lucky
- 7 Solution 6 - Overkill
- 8 🎥✨ Video Solution by Scholars Foundation ➡️ Easy-to-Understand 💡✔️
- 9 Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)
- 10 Video Solution 2 by SpreadTheMathLove
- 11 Video Solution by Daily Dose of Math
- 12 See also
Problem
A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
Solution 1 - Prime Factorization
We can start by assigning the values x and y for both sides. Here is the equation representing the area:
Let's write out 2024 fully factorized.
Since we know that , we want the two closest numbers possible. After some quick analysis, those two numbers are and .
Now we multiply by and get
Solution by IshikaSaini.
Solution 2 - Squared Numbers Trick
We know that . Recall that .
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 we get
Solution by ~Taha Jazaeri
Note: The square of any positive integer with units digit , written in the format where is a positive integer, is equal to .
~Cattycute
Solution 3 - AM-GM Inequality
Denote the numbers as . We know that per AM-GM, , but since , must be slightly less than 90, so 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
-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.
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:
~BenjaminDong01
Solution 5 - Get Lucky
Note: This is what I did.
Assuming it's a square,
~BenjaminDong01
Solution 6 - Overkill
Let denote the length and denote the width of the rectangle. We have and we wish to minimize . WLOG, let
From , we have that . So we minimize the expression . The derivative of this expression is We set the derivative equal to zero to find the absolute extrema of . In other words, Solving this yields , which means the perimeter has a maximum when has this value. Only the positive value of is feasible since we deal with side lengths. However, we also want to be an integer.
We claim that is a monotonically increasing function when . Looking back at the function of the derivative, , it is positive when , which means that is monotonically increasing on this domain.
Thus, we want to be the greatest integer less than to maximize the perimeter and satisfy the problem's constraints. So, and , and the desired perimeter is .
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
~Thesmartgreekmathdude
See also
2024 AMC 10B (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.