SOLUTION: Diana has available 400 yards of fencing and wishes to enclose a rectangular area.
(a) Express the area A of the rectangle as a function of the width W of the rectangle.
(b
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(a) Express the area A of the rectangle as a function of the width W of the rectangle.
(b
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Question 1138949: Diana has available 400 yards of fencing and wishes to enclose a rectangular area.
(a) Express the area A of the rectangle as a function of the width W of the rectangle.
(b) For what value of W is the area largest?
(c) What is the maximum area? Found 2 solutions by Boreal, ikleyn:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! Draw this
sides are x x and 400-2x
The area is x(400-2x)=-2x^2+400x
the vertex for this quadratic is a maximum and will have x value of -b/2a=-400/-4=100 yds
the maximum area for this 100*200 rectangle is 20000 yds^2.
The formula for A=-2W2+400W for W=width
The post by @boreal represents the solution in a very inaccurately way, and is incorrect in many places.
I came to bring the correct solution.
sides are x and 200-x (where 200 = 400/2 is half of the perimeter)
The area is x(200-x) = -x^2 + 200x
the vertex for this quadratic is a maximum and will have x value of
-b/2a = (-200)/(-2) = 100 yds
The maximum area for this 100*100 SQUARE (!) is 10000 yds^2.
The formula for the area is = -W^2 + 200*W for W= width.
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The answer is very well known fact: under given condition, the maximum area is provided by the square,
and when the perimeter of a rectangle is given (= the fence length), the side of this square is one fourth of the perimeter.
