Question 4627
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 {{{Y = ax^2 + bx + c }}}, 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


{{{Y = ax^2 + bx + c }}}, 
Substituting into this equation YIELDS [Pun!!  Get it??]:
{{{8750 = a(5)^2 + b(5) + c}}}
{{{15000 = a(10)^2 + b(10) + c }}}
{{{ 0 = a(0)^2 + b(0) + 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^2 + 2000x }}}, which is a parabolic graph opening down.


The maximum yield occurs at the vertex, which can be found at 
{{{x = -b/2a}}}

{{{x = -2000/(2*(-50)) = -2000/(-100) = 20}}}
{{{Y(20) = -50*20^2 + 2000*20 = -20000 + 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