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{(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> inches. |
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| + | == See Also== | ||
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| + | {{AMC10 box|year=2011|ab=B|num-b=2|num-a=4}} | ||
Revision as of 15:42, 4 June 2011
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 
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 | ||
