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Difference between revisions of "2021 AMC 10B Problems/Problem 11"

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~Bryguy
 
~Bryguy
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== Video Solution by OmegaLearn (Simon's Favorite Factoring Trick) ==
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https://youtu.be/vWlRQiyt0c8
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2021|ab=B|num-b=10|num-a=12}}
 
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{{MAA Notice}}

Revision as of 02:32, 12 February 2021

Problem

Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?

$\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64$

Solution 1:

Let the side lengths of this rectangular pan be $m$ and $n$; it follows that $(m-2)(n-2) = \frac{mn}{2}$. This gives $(m-4)(n-4) = 8$ after some manipulation, so $(m, n) = (5, 12), (6, 8)$. By inspection, $5 \cdot 12 = \boxed{\textbf{(D) }60}$ maximizes the number of brownies ~ ike.chen

Solution 2:

Let the dimensions of the rectangular pan be $x$ and $y$. The number of interior pieces is $(x-2)(y-2)$ because you cannot include the border, and the number of pieces along the perimeter is $\frac{xy}{2}$ (THIS PART IS FLAWED ~ anonymous user).

Setting these two expressions equal, we have $\frac{xy}{2}=(x-2)(y-2) \Rightarrow xy=2(xy-2y-2x+4) \Rightarrow xy-4x-4y+8=0$

Applying SFFT (Simon's Favorite Factoring Trick), we get $(x-4)(y-4)=8$. Doing a bit of trial-and-error, we see that $xy$ is maximum when $x=5$ and $y=12$, which gives us a maximum of $60$ brownies. $\Rightarrow \boxed{\textbf{(D) }60}$.

~Bryguy


Video Solution by OmegaLearn (Simon's Favorite Factoring Trick)

https://youtu.be/vWlRQiyt0c8

~ pi_is_3.14

See Also

2021 AMC 10B (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 10 Problems and Solutions

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