2011 AMC 10B Problems/Problem 3

Problem

At a store, when a length is reported as $x$ inches that means the length is at least $x - 0.5$ inches and at most $x + 0.5$ inches. Suppose the dimensions of a rectangular tile are reported as $2$ inches by $3$ inches. In square inches, what is the minimum area for the rectangle?

$\textbf{(A)}\ 3.75 \qquad\textbf{(B)}\ 4.5 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8.75$

Solution 1

The minimum dimensions of the rectangle are $1.5$ inches by $2.5$ inches. The minimum area is $1.5\times2.5=\boxed{\mathrm{(A) \ } 3.75}$ square inches.

Solution 2 (Utilize algebra)

Notice that the minimum area of a rectangle with dimensions $x$ inches by $y$ inches will result in $(x - 0.5)\times(y - 0.5)$ . Simplify to get $xy - 0.5x - 0.5y + 0.25$ . Plug in $2$ and $3$ for $x$ and $y$ to get $6 - 1 - 1.5 + 0.25 = \boxed{\text{(A)\ 3.75}}$ square inches.

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 10 Problems and Solutions

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2011 AMC 10B Problems/Problem 3

Contents

Problem

Solution 1

Solution 2 (Utilize algebra)

See Also