
A police officer seals off the scene of an accident using a roll of yellow tape that is 376 feet long. What dimensions should be used to seal off the maximum rectangular area around the collision? (Enter your answers as a commaseparated list.) Find the maximum area.
let
x = the length of the rectangle
y = the width of the rectangle
x > 0, y > 0, x >= y
the perimeter of rectangle P = 2*(length + width) that is
P = 2*(x + y)
and since the tape is 376 ft long we have P = 376 that is
2*(x + y) = 376
x + y = 188 => y = 188 x
the area of a rectangle is given my the formula A = length * width
in our case A = x*y = x (188  x)=  x^2  188x
A(x) =  x^2  188x
we need to find the maximum of the quadratic function A(x) =  x^2  188x
A(x) =  x^2 + 188x is parabolic function with quotient 1 in front of x^2 thus it has maximum in the vertex
A max = c  b^2/4a
A max =  188^2 / (4)
A max = 8836
the maximum area is 8836 ft^2
if we solve  x^2 + 188x = 8836 we find x = 94 ft
y = 188  94 = 94 ft
the dimensions of the rectangle with maximum area are length of 94 ft and width of 94 ft that is the rectangle is a square.
the maximum area is 8836 ft^2.
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