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2011 AMC 10B Problem 3

2011 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 1 of 5 · about 1 min · position P1-10

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?

A)
$3.75$
B)
$4.5$
C)
$5$
D)
$6$
E)
$8.75$

Answer

Key insight

Both dimensions are positive, so the smallest area comes from the smallest possible length and width together: 1.5 times 2.5.

Solution

A reported $2$ inches means the true width is between $1.5$ and $2.5$ inches, and a reported $3$ inches means the true length is between $2.5$ and $3.5$ inches.

The area is the product of two positive quantities, so it is smallest when both are as small as allowed:
$1.5 \times 2.5 = 3.75.$

The minimum area is $\boxed{\textbf{(A)}\ 3.75}$ .

Why this works

For positive quantities, shrinking either factor shrinks the product, so the extreme of a product is reached at the extremes of the factors. Interpreting "reported as $x$ " as the interval $[x - 0.5, x + 0.5]$ is the only reading step.

The trap

Using the reported dimensions 2 by 3 (area 6) or shrinking only one dimension, giving 4.5 or 5.

Common mistakes

Using the reported dimensions $2$ by $3$ (area $6$ ) or shrinking only one dimension, giving $4.5$ or $5$ .
Taking the maximum instead of the minimum ( $2.5 \times 3.5 = 8.75$ ).

Techniques

Consider the largest/smallest element or boundary case