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.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! The length and width (measured in cm) are whole numbers.
When you multiply one times the other, the product is 48 (because that would be the area, 48 square centimeters).
Ask your 5th grader for pairs of numbers whose product is 48.
You could list them:
1 and 48
2 and 24
3 and 16
4 and 12
6 and 8
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).
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.
Let your fifth grader calculate all he/she wants, until convinced of the answer.
NOTES:
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.
This kind of problem, with no restrictions on the dimensions is used for algebra problems, and we could make it into calculus problems too.
|
|
|
