seal off a rectangular scene of an accident using a roll of yellow tape that is 80ft

Question

a police officer wants to seal off a rectangular scene of an accident using a roll of yellow tape that is 80ft long. The accident occurred along the side of a wall therefore the officer does not need to put any yellow tape along the wall. what is the largest possible area that the officer can enclose and what are the dimensions of the enclosed area?

Answer 1

Hi Cole:

Let's start by using some ASCII art to draw a picture of the situation.

–––––––––––––––––––––––––––––––––––––––––––––––––

| |

w | |

|_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ |

l

In the picture above, the long solid line represents the wall they discuss in the problem description, and the dashed lines represent the area we're going to cordon off with our 80 ft of yellow tape.

Now let's write some equations for the area A and the perimeter P.

1) A = lw

2) P = l + 2w = 80

The area of a rectangle is simply length times width, and the perimeter in this case is just the length of the one side for which we use the yellow tape (the other side is the wall) plus twice the width for which we use the yellow tape. (I've dropped the unit on the 80 ft for now, but we'll pick it back up later on).

Now, since we want to maximize the area, we want an equation in one variable, but currently our equation for the area is in terms of both l and w. So, we'll use the second equation to solve for one of the variables in terms of the other. This is ultimately arbitrary, but I think I like better solving for l.

l = 80 - 2w

And we put this information into the first equation so that we have the area as a function of width only:

A(w) = (80 - 2w)w = 80w - 2w²

Now we can maximize, taking the derivative and setting it equal to zero.

dA/dw = 80 - 4w = 0 ⇒ w = 20

So, the width is 20 ft, and the length (back substituting into our equation l = 80 - 2w) is 40 ft.

Thus, the maximum area we can cordon off with this 80 ft of yellow tape is (40 ft)(20 ft) = 800 ft², and the dimensions of the rectangular area are l = 40 ft, and w = 20 ft.

Hope this helps!

-Joel