SOLUTION: The perimeter of a rectangle remains fixed at 16cm. By first writing, an equation involving the sides of the rectangle, find the dimensions that give the maximum area.
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Question 1120518: The perimeter of a rectangle remains fixed at 16cm. By first writing, an equation involving the sides of the rectangle, find the dimensions that give the maximum area. Found 2 solutions by Boreal, ikleyn:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! Sides are x and 8-x, to make a half perimeter of 8. 2x+16-2x=16
Maximize the product of those 2
x(8-x)=-x^2+8x
Using calculus, first derivative is -2x+8=0 or 2x=8 and x=4, a square 4 x 4 with area 16
Or, parabola with vertex x value at -b/2a which occurs at -8/-2 or 4, and the same result.
You can put this solution on YOUR website! .
At given perimeter P of a rectangle, the greatest area is achieved when the rectangle is a square
(and then its side is, obviously, P/4 units of length).
