Difference between revisions of "2011 AMC 10B Problems/Problem 3"
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== Solution == | == Solution == | ||
− | The minimum dimensions of the rectangle are <math>1.5</math> inches by <math>2.5</math> inches. The minimum area is <math>1.5\times2.5=\boxed{\mathrm{(A) \ } 3.75}</math> inches. | + | The minimum dimensions of the rectangle are <math>1.5</math> inches by <math>2.5</math> inches. The minimum area is <math>1.5\times2.5=\boxed{\mathrm{(A) \ } 3.75}</math> square inches. |
== See Also== | == See Also== |
Revision as of 21:30, 16 January 2016
Problem
At a store, when a length is reported as inches that means the length is at least inches and at most inches. Suppose the dimensions of a rectangular tile are reported as inches by inches. In square inches, what is the minimum area for the rectangle?
Solution
The minimum dimensions of the rectangle are inches by inches. The minimum area is square inches.
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.