SOLUTION: 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,

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: 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,      Log On

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Question 4658: 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 hs yield will vary quadratically with the nnumber 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 had no production when he was grazing no cattle.
a) Define the variables, write the ordered pairs, and find the particular equation of this function.
b) What number of cattle will gie him his maximum profit in number of punds?
c)What is this weight?
d) What is the domain and range of this function?
Answer by rapaljer(4667) About Me  (Show Source):
You can put this solution on YOUR website!
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

Y+=+ax%5E2+%2B+bx+%2B+c+,
Substituting into this equation YIELDS [Pun!! Get it??]:
8750+=+a%285%29%5E2+%2B+b%285%29+%2B+c
15000+=+a%2810%29%5E2+%2B+b%2810%29+%2B+c+
+0+=+a%280%29%5E2+%2B+b%280%29+%2B+c+

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 Y+=+-50x%5E2+%2B+2000x+, which is a parabolic graph opening down.

The maximum yield occurs at the vertex, which can be found at
x+=+-b%2F2a
x+=+-2000%2F%282%2A%28-50%29%29+=+-2000%2F%28-100%29+=+20
Y%2820%29+=+-50%2A20%5E2+%2B+2000%2A20+=+-20000+%2B+40000+=+20000 = 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