Question 694549
The city of Ottawa has provided 300 meters of rope to enclose a rectangular swimming area along the shore of the Mooney Bay Beach. What is the maximum area that can be enclosed and the dimensions of the maximum area. 
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Let L = the length of the rectangle
Let W = the width
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Only 3 sides are required so the perimeter with 300 m of rope will be:
L + 2W = 300
L = (300-2W); use this form for substitution in the area equation
:
A = L * W
Replace L with (300-2W)
A = W(300-2W)
A = -2W^2 + 300W
this is a quadratic equation, max area occurs at the axis of symmetry which we can find with formula x = -b/(2a), in this equation it's
W = {{{(-300)/(2*-2)}}}
W = {{{(-300)/-4)}}}
W = +75 m is the width for max area
then
L = 300 - 2(75)
L = 300 - 150
L = 150 m is the length
Find the area
150 * 75 = 11,250 sq/m is the max area with 300 ft of rope
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How about this? Are you understanding this now?