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

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

Revision as of 21:45, 11 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

Let the dimensions of the rectangular 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 $\frac{xy}{2}$.

Setting these two 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. $\Rightarrow \boxed{\textbf{(D) }60}$.

Solution by Bryguy

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