Answer:
<h2>C. <em>
20,160</em></h2>
Step-by-step explanation:
This question bothers on permutation since we are to select a some people out of a group of people and then arrange in a straight line. If r object are to be arranged in a straight line when selecting them from n pool of objects. This can be done in nPr number of ways.
nPr = n!/(n-r)!
Selection of 6 people out of 8 people can therefore be done in 8C6 number of ways.
8P6 = 8!/(8-6)!
8P6 = 8!/2!
8P6 = 8*7*6*5*4*3*2!/2!
8P6 = 8*7*6*5*4*3
8P6 = 56*360
8P6 = 20,160
<em>Hence this can be done in 20,160 number of ways</em>
Answer:
- 12 ft parallel to the river
- 6 ft perpendicular to the river
Step-by-step explanation:
The least fence is used when half the total fence is parallel to the river. That is, the shape of the rectangle is twice as long as it is wide.
72 = W(2W)
36 = W²
6 = W . . . . . . the width perpendicular to the river
12 = 2W . . . . the length parallel to the river
_____
<em>Development of this relation</em>
Let T represent the total length of the fence for some area A. Then if x is the length along the river, the width is y=(T-x)/2, and the area is ...
A = xy = x(T -x)/2
Note that the equation for area is that of a parabola with zeros at x=0 and at x=T. That is, for some fence length T, the area will be a maximum at the vertex of this parabola. That vertex is located halfway between the zeros, at ...
x = (0 +T)/2 = T/2
The corresponding area width (y) is ...
y = (T -T/2)/2 = T/4
Equivalently, the fence length T will be a minimum for some area A when x=T/2 and y=T/4. This is the result we used above.
