Question

A lifeguard needs to rope off a rectangular swimming area in front of long lake beach, using 2500 yd of rope and floats. What dimensions of the rectangle will maximize the area? What is the maximum area? (Note that the shoreline is one side of the rectangle.)

Answer 1

Answer:

Dimensions :

x = 625 yd

y = 1250 yd

A (max)  = 781250 yd²

Step-by-step explanation:

Let "x"  be the small side of the rectangle,  and "y" the longer

A = x*y       perimeter  is  2500 yd  = 2x + y

Then  y = 2500 - 2x     (1)

A(x) = x * ( 2500 - 2x )        ⇒  A(x) = 2500x  - 2x²

Taking derivatives :

A´(x)  =  2500 - 4x              and     A´(x)  = 0

2500 - 4x  = 0      4x = 2500     x = 2500/4

x = 625 yd

Now by substtution of x value in equatio (1)

y =  2500 - 2x       ⇒  y = 2500 - 2* 625

y = 2500 - 1250

y = 1250 yd

And fnally th aea is:

A (max) = 1250 * 625

A (max) = 781250 yd²

Answer 2

For maximum area, the rectangle should have a length of 625 yard and a width of 625 yard

A rectangle is a quadrilateral (has four sides and four angles) in which opposite sides are parallel and equal to each other. Also all the angles of a rectangle measure 90 degrees each.

Let x represent the length and y represent the width of the swimming area.

Since 2500 yd of rope and floats is available, hence:

Perimeter = 2(x + y)

2500 = 2(x + y)

x + y = 1250

y = 1250 - x

Area of a rectangle = length * breadth

Area(A) = xy

A = x(1250 - x)

A = 1250x - x²

The maximum area is at dA/dx = 0

dA/dx = 1250 - 2x

2x = 1250

x = 625 yard

y = 1250 - x = 1250 - 625 = 625 yard

Hence for maximum area, the rectangle should have a length of 625 yard and a width of 625 yard

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