Algebra question.

Question

A summer camp has 520 ft of float line with which to rope off three adjacent rectangular areas of a lake for swimming​ lessons, one for each of three levels of swimming ability. A beach forms one side of the swimming areas. Suppose that the width of each area is x feet. Answer parts​ a) through

​c).

a) Express the total area of the three swimming areas as a function of x.

A(x)=

Can't do B or C until I figure out question A.

Answer 1

Hello Britney,

I did a quick drawing to illustrate the word problem. The bottom length of the rectangle is the beach, so you won't need float line for it.

As the drawing shows, you will use float line for 4 different widths, all equal to x. Since you're starting off with 520 feet, that leaves you with 520 - x - x - x -x which is 520 - 4x that will be used for the top length.

Area is Length * width, so in this case the total area will be: A(x) = x * (520 - 4x) which you could expand to A(x) = 520x - 4x²

Cheers.

Answer 2

520-4x = the length of the total area, draw a diagram

x = width

Area = LW = length times width = [(520-4x)](x) = 540x-4x^2

A(x) = 520x -4x^2

A =4x(130 -x)

0<x<130

520/4 = 130. If x= 130, the length would = 0,

x has to be less than 130.

if x= 0 or less, the area = 0 or is negative

the domain is (0, 130)

another question is what dimensions would maximize the total area of the roped in areas?

A = 520x - 4x^2

take the derivative and set equal to zero, solve for x

A' = 520 - 8x = 0

8x = 520

x = 520/8 = 64 feet = area maximizing width

area maximizing length = 520 -4(64) = 520-256 = 264 feet

maximum area = 64 x 264 = 16,896 square feet for the total area

maximum area for each of the 3 subareas = 16,896/3 = 5,632 ft^2

each of the 3 areas, with max area, has dimensions 64 by 264/3

= 64 ft by 88 ft

64(88) = 5632

3(5632) = 16,896 ft^2 = 64(264)

range is (0, 16,896)