SOLUTION: A dog owner has 250ft of fencing to enclose a rectangular run area for his dog. If he wants the maximun possible ares, what should the length and width of the rectangle be?
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Question 421354: A dog owner has 250ft of fencing to enclose a rectangular run area for his dog. If he wants the maximun possible ares, what should the length and width of the rectangle be?
i did 2x+2z=250 then got the equation 2z=250-2x using -b/2a i got -250/-4=62.5=x
then 2z=250-2(62.5) and got z =62.5 which would mean each side is 62.5 and that checks out because 62.5*4=250 but how do i know thats the maximun area and also thats a square since all sides are equal?
can anyone please please tell me what i did wrong or what i need to do?
thank you!!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A dog owner has 250ft of fencing to enclose a rectangular run area for his dog. If he wants the maximun possible ares, what should the length and width of the rectangle be?
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The max area for a rectangle is a square.
The perimeter = 250 ft = 2L + 2W
--> L + W = 125
W = 125 - L
--------
Area = L*W = L*(125 - L)
is a parabola. The max is at the vertex
The line of symmetry is L = -b/2a
L = 125/2 = 62.5
The vertex is on the line of symmetry at f(-b/2a) = f(62.5)
That proves that 62.5 ft gives the max area for the given perimeter.
Max area = 62.5^2 = 3906.25 sq ft
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