Question 75650: Enclosing a field: You have 16 miles of fence that you will use to enclose a rectangular field.
a. Draw a picture to show that you can arrange the 16 miles of fence into a rectangle of width 3 miles and length 5 miles. What is the area of the rectangle. (not wanting a picture, but just the area)
b. Draw a picture to show that you can arrange the 16 miles of fence into a rectangle of width 2 miles and length 6 miles. What is the area of the rectangle?
C. The first two parts of this problem are designed to show you that you can get different areas for rectangles of the same perimeter, 16 miles. In general if you arrange the 16 miles of fence into a rectangle width w miles, then it will enclose an area of A=w(8-W) square miles.
1. Make a graph of the area enclosed as a function of w, and explain what the graph is showing.
2. What width w should you use to enclose the most area?
3. What is the length of the maximun area rectangle that you have made, and what kind of figure do you have?
Answer by renevencer22(21)   (Show Source):
You can put this solution on YOUR website! a) L = 5
W = 3
perimeter = 2*5 + 2*3 = 16 miles
area = 3*5 = 15 sq. miles
b) L = 6
W = 2
perimeter = 2*6 + 2*2 = 16 miles
area = 2*6= 12 sq. miles
c)
1. see figure the graph. Shows area as function of width:
a second degree, a = w*(8-w); graph ends at width = 4 mi.
Pls see: use copy, paste (to your file) and ctrl+click
http://www.freewebs.com/renevencer11/-%20math/areavswidth.jpg
http://www.freewebs.com/renevencer11/-%20math/table01.jpg
2. for maximum area: w = 4 mi
3. for maximum area: length = 4 mi. The result is a square.
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