document.write( "Question 583272: Find the whole number dimensions of a rectangle that has the given area and the least perimeter possible. 48 cm2 I want to know the best way to explain this to my 5th grader. \n" ); document.write( "

Algebra.Com's Answer #372397 by KMST(5328) $\"\"$ $\"About$

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The length and width (measured in cm) are whole numbers.
\n" ); document.write( "When you multiply one times the other, the product is 48 (because that would be the area, 48 square centimeters).
\n" ); document.write( "Ask your 5th grader for pairs of numbers whose product is 48.
\n" ); document.write( "You could list them:
\n" ); document.write( "1 and 48
\n" ); document.write( "2 and 24
\n" ); document.write( "3 and 16
\n" ); document.write( "4 and 12
\n" ); document.write( "6 and 8
\n" ); document.write( "The perimeter would be two width plus two lengths. You can add a width and a length first, and then double the sum. (I like that way of calculating perimeters better, because I can do that in my head).
\n" ); document.write( "When you look at the pairs above, you realize that 6+8=14 for a perimeter of 28 cm is the smallest sum. The rectangle is 6 cm wide and 8 cm long.
\n" ); document.write( "Let your fifth grader calculate all he/she wants, until convinced of the answer.
\n" ); document.write( "NOTES:
\n" ); document.write( "The largest area for a given perimeter, or the shortest perimeter for a given area means a circle. Because round pens or garden enclosures are not practical, we prefer rectangles. The best rectangle for cost of fencing is a square. In this case, because we wanted whole number measurements we end up with the \"squarest\" rectangle, the closest pair of factors.
\n" ); document.write( "This kind of problem, with no restrictions on the dimensions is used for algebra problems, and we could make it into calculus problems too. \n" ); document.write( "

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