PART 1 – Maximum Space To model the following fencing problem

Question

- If the last two digits are less than 50, add them to 50 (ie. 35 + 50 = 85 m of fencing)

- If your last two digits are greater than or equal to 50, use that number. (ie. 87 = 87 m of fencing)

What is your student number ? ______668997_______________

What length of fencing will you work with? ______97__________m

You want to create an enclosure with three equal sections (see Fig 1).

What are the dimensions of the largest possible rectangular enclosure that you can create using your fencing?

What is the domain and range of the variables you’ve chosen? Show all your work. (i.e. Be sure to state let statements for your variables and use appropriate conventions like function notation. Also, you must use mathematical modelling, which means that you must set up an equation that you then solve to lead to the answer, not just use guess and check.)

Please leave your final answer in exact form

https://docs.google.com/document/d/1aJrMZrDFqqBTssS_JxGJSJTx5-P3wDBP9d5037DxHPg/edit?usp=sharing

Answer 1

By largest rectangle means rectangle with the largest area given the total fence length.

Let L = the length of the rectangle and W = the width of the rectangle. The width will be divided into 3 sections which means two extra divider fences.

Perimeter plus inner fencing, P for a rectangle is 4L + 2W= 97.

L = (97-2W)/4 = 24.25 - W/2

The total area, A = LW = 24.25W-W²/2. To maximize the area, take the derivative of the area, set that to zero and solve for W.

A(W) = 24.25W-W²/2

A'(W)= 24.25-W

W= 24.25 m

L = 24.25-24.25/2 = 12.125 m