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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 comma-separated 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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