Question 4627: A rancher wants to yield the maximum amount of edible beef off his square mile of land. At first he finds that as he adds more cattle, his yield goes up. However, if he overgrazes, the harvest goes down. An agricultural agent tells him that his yield will vary quadratically with the number of animals that he grazes. For five head of cattle, his production is 8,750 pounds of beef, and for 10 head, he reaps 15,000 pounds of beef. He has no production when he was grazing no cattle.
a) Define the variables, write the odered pairs, and find the particular equation of this function.
b) What number of cattle will give him his maximum profit in number of pounds?
c) What is this weight?
d) What is the domain and range of this function?
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! I think we need Longjonsilver or Kenny for this one, but I'll give it a try. First I had never heard of varying quadratically before, but I'm assuming that it means  , where x is the number of animals, and Y is the Yield in pounds of beef.
Three pieces of information are given in the problem to allow you to solve for the constants a, b, and c.
Y(5) = 8750
Y(10) = 15000
Y(0) = 0
,
Substituting into this equation YIELDS [Pun!! Get it??]:
8500 = 25a + 5b
15000= 100a + 10b
0 = 0(a) + 0(b) + c
The last equation gives you immediately that c = 0.
Multiply the first equation by -2:
-17500 = -50a - 10b
15000 = 100a + 10b
-2500 = 50a
a=-50
Substituting back into either equation gives you b = 2000
The equation of Yield is  , which is a parabolic graph opening down.
The maximum yield occurs at the vertex, which can be found at
 = maximum yield
Domain of this problem is from x = 0 to 40: [0, 40]
Range Y is from 0 to the maximum at 20,000: [0, 20,000]
R^2 at SCC
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