Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
 
<math>\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64</math>
 
<math>\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64</math>
  
−
==Solution==
+
==Solution 1:==
−
Let the dimensions of the rectangular be <math>x</math> and <math>y</math>. The number of interior pieces is <math>(x-2)(y-2)</math> (because you cannot include the border) and the number of pieces along the perimeter is <math>\frac{xy}{2}</math>.
+
Let the side lengths of this rectangular pan be <math>m</math> and <math>n</math>; it follows that <math>(m-2)(n-2) = \frac{mn}{2}</math>. This gives <math>(m-4)(n-4) = 8</math> after some manipulation, so <math>(m, n) = (5, 12), (6, 8)</math>. By inspection, <math>5 \cdot 12 = \boxed{\textbf{(D) }60}</math> maximizes the number of brownies ~ ike.chen
  
−
Setting these two equal, we have <math>\frac{xy}{2}=(x-2)(y-2) \Rightarrow xy=2(xy-2y-2x+4) \Rightarrow xy-4x-4y+8=0</math>
+
==Solution 2:==
 +
Let the dimensions of the rectangular pan be <math>x</math> and <math>y</math>. The number of interior pieces is <math>(x-2)(y-2)</math> because you cannot include the border, and the number of pieces along the perimeter is <math>\frac{xy}{2}</math> (THIS PART IS FLAWED ~ anonymous user).
 +
 
 +
Setting these two expressions equal, we have <math>\frac{xy}{2}=(x-2)(y-2) \Rightarrow xy=2(xy-2y-2x+4) \Rightarrow xy-4x-4y+8=0</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>.
 
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>.
  
−
Solution by Bryguy
+
~Bryguy

Revision as of 23:14, 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 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_11&oldid=145737"