I need to build a rectangular enclosure for my pet pig. I only have 36 yards of fencing to use to build this rectangular pen. Perimeter Question!!

Question

1) Using the fact that P=2l +2w, express the length L in terms of W.

2) Now express the Area of the rectangle as a function of the width of the rectangle.

3) If I use only 4 yards of fencing for the width of my rectangle, what is the area.

4) What should the dimensions of my rectangle by in order to have the biggest area for Izzy to live in? (You should have a quadratic function and figure out where the maximum occurs) What will be the area?

Answer 1

1) P = 36 = 2l+2w

 

2) The first step is to rearrange the formula for perimeter and solve for l in which is

 

l = (36-2w)/2 = 18-w

 

Now the area of the rectangle is:

 

A=lw=(18-w)w = 18w-w² 

 

3) if you use 4 yds of fencing for the width of the rectangle, the area of your fence is:

 

A=18*4-4² = 72-16=56 yds²

 

4) The way to find the maximum area in which lzzy could live in is to complete the square.

 

The first step is to factor out -1 from the equation in which is:

 

A = -1(w² - 18w) = -1(w² - 18w +81) + 81

 

A = -1(w - 9)² +81

 

Hence, the vertex of the parabola occurs at w = 9 yds and A = 81 yds².

 

This implies l=18-9=9 yds. Therefore, the rectangular enclosure must be a square with length of side being 9 yds.