# 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? i di

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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(30937)   (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? ---------------- 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