Question 1675: Question regarding quadradic equations in intermediate algebra:
Determine the dimensions of a rectangle with a perimeter of 20 that encloses the greatest area. The perimeter is the sum of all sides of the rectangle. Is there another shape that would be more efficient (higher area to perimeter ratio?)
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Let te dimensions of the rectangles be x,y.
Since the perimeter is 2(x+y)= 20,so x+y = 10.
The area A = xy = x(10-x) = -x^2 + 10x
(by complete square)
= -(x^2 - 10x + 25) + 25
= -(x - 5)^2 + 25
(since -(x - 5)^2 <= 0 for all x)
A has max value if x =5 and so y = 10-5 =5.
Hence, when the rectangle dimensions are 5, 5 , its area is max.
Also, note that in this case, Area/perimeter = 25/20 = 1.25
If the shape is a circle, then we see that
Area/perimeter = (pi R^2)/(2pi R) = R/2
Therefore, when the radius R > 2.5 , the circle is more efficient
than the rectangle. (higher area to perimeter ratio)
Kenny
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