SOLUTION: how do i find the quadratic equation,with this question:
" A rancher wants to yield the max amount of edible beef off of his square mile of land. At first he finds that as he adds
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Question 251715: how do i find the quadratic equation,with this question:
" A rancher wants to yield the max amount of edible beef off of his square mile of land. At first he finds that as he adds more cattle, his yield goes up. However, if he overgrazes, the yield goes down. An agricultural agent tell him that his yield will vary quadratically with the # of animals that he grazes. For 5 head of cattle, his production is 8750 pounds of beef, and for 10 head of cattle, he reaps 15,000 pounds of beef. He had no production when he was grazing no cattle."
Find the particular equation of this function expressing pounds of beef produced in terms of the number of cattle.
Found 2 solutions by ankor@dixie-net.com, Theo:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
" A rancher wants to yield the max amount of edible beef off of his square mile of land. At first he finds that as he adds more cattle, his yield goes up.
However, if he overgrazes, the yield goes down. An agricultural agent tell him that his yield will vary quadratically with the # of animals that he grazes.
For 5 head of cattle, his production is 8750 pounds of beef, and for 10 head of cattle, he reaps 15,000 pounds of beef.
He had no production when he was grazing no cattle."
Find the particular equation of this function expressing pounds of beef produced in terms of the number of cattle.
:
Using the form; ax^2 + bx + c = y, c=0, find a and b
x=5 (no. of cattle); y=8750 (lb of beef)
a(5^2) + 5b = 8750
25a + 5b = 8750
and
x=10; y=15000
100a + 10b = 15000
:
Multiply the 1st equation by 2, subtract from the above equation
100a + 10b = 15000
50a + 10b = 17500
---------------------subtraction eliminates b, find a
50a = -2500
a =
a = -50
;
;
Find a using the 1st equation, replace a with -50
25a + 5b = 8750
25(-50) + 5b = 8750
-1250 = 5b = 8750
5b = 8750 + 1250
5b = 10000
b =
b = 2000
:
The quadratic function (lb of beef for No. of cattle): f(x) = -50x^2 + 2000x
;
:
Check solution using x=10
y = -50(10^2) + 2000(10)
y = -50(100) + 20000
y = -5000 + 20000
y = 15000 as given in the problem
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
standard form of a quadratic equation is:
y = f(x) = ax^2 + bx + c = 0
a = coefficient of x^2 term.
b = coefficient of x term.
c = constant term
x represents head of cattle.
with 0 head of cattle, he gets no yield so the first equation you have is:
y = f(0) = a*0^2 + b*0 + c = 0
this tells him that the constant term of c has to be equal to 0 **********
with 5 head of cattle, he gets 8750 pounds of beef.
this makes the standard form of the quadratic equation equal to:
y = f(5) = a*5^2 + b*5 = 8750
with 10 head of cattle, he gets 15000 pounds of beef.
this makes the standard form of the quadratic equation equal to:
y = f(10) = a*100^2 + b*10 = 15000
he needs to solve for the coefficients of the quadratic equation.
the 2 equations become:
y = f(5) = 25*a + 5*b = 8750 (equation 1)
y = f(10) = 100*a + 10*b = 15000 (equation 2)
he needs to solve these two equations simultaneously to get the value of a and b.
he already has the value of c = 0.
solving these equations simultaneously, he gets;
a = -50
b = 2000
his quadratic equation becomes:
y = f(x) = -50x^2 + 2000x
when x = 5, this equation becomes:
y = f(5) = -50*25 + 2000*5 = -1250 + 1000 = 8750
when x = 10, this equation becomes:
y = f(10) = -50*100 + 2000*10 = -5000 + 20000 = 15000
the values are good.
since the coefficient of the x^2 term is negative, the parabola formed by this equation will point upward and open downward.
it will peak when x = -b/2a.
That is the maximum point of the equation.
x = -b/2a becomes x = -2000/-100 = 20
when x = 20, the number of pounds of beef he produces will be:
y = f(20) = -50*20^2 + 2000*20 = -50*400 + 40000 = -20000 + 40000 = 20000.
his production of beef will peak at 20000 when he has 20 head of cattle.
graph of the his equation is shown below:
the horizontal line is at y = 20000 pounds of beef.
if you need help with solving the equations simultaneously, let me know and I'll send you the solution for that.
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