Cattle Grazing –Key
To give cattle room to graze, a farmer decides to build a fence in the shape of a
rectangle. He has 2100 feet of fencing available.
A. If the width of one side of the fence is 400 feet, what would the other
dimension have to be?
2100 – 2(400) = 300 left for 2 sides. Divide by 2 to get 150 feet for the
other dimension.
B. The theoretical minimum value for one of the sides is 0 feet. What is the
theoretical maximum value for one of the sides?
2100/2 = 1050 ft
C. Complete table by finding the length given a width. Then determine the
Area for each set of values.
Width
0
100
200
300
400
500
600
700
800
900
1000
1100
Length
1050
950
850
750
650
550
450
350
250
150
50
NA
Area (ft2)
0
95000
170000
225000
260000
275000
270000
245000
200000
135000
50000
NA
D. What conclusions can you reach regarding the dimensions and the area?
Different dimensions yield a different areas.
E. Use the TI-83/84 or Excel to investigate how changing the dimensions of
the rectangle affects the rectangle’s area.
In Excel they could set the increment to any value (100 is shown below)
Extension 1.
Suppose you wanted to build the fence alongside a barn. That is, the barn wall
will serve as one of the sides of the fence. Assume the farmer still has 2100 ft of
fencing. Use the TI-83/84 or excel to determine the maximum area. (Hint: Use
increments of 25).
Note: It may be easier to just
put in x(2100 – 2x) into Y1.
Answer: So the dimensions would be 525 ft by 1050 ft for a maximum area of
551,250 ft2
Extension 2.
Suppose that the farmer wanted to place a partition down the middle of the fence
and not use the barn wall as a fence side. Assume the farmer still has 2100 ft of
fencing. Use the TI-83/84 or excel to determine the maximum area.
The actual Max occurs when x = 16.67. You can use this to discuss the limitations
of using the table.