q = 8000 - 400p
q = the numbe of gallons of milk sold each day.
p = the price per gallon.
revenue equals quantity * price.
if r = revenue and q = quantity and p = price, then the formula becomes:
r = q * p
since q = 8000 - 400p, you can replace q in that equation with it's equivalent value in terms of p to get:
r = (8000 - 400p) * p
simplify this equation to get:
r = 8000p - 400p^2.
when revenue = 40,000, the equation becomes:
40000 = 8000p - 400p^2.
add 400p^2 to both sides of this equation and subtract 8000p from both sides of this equation and order the terms in descending order of degree to get:
400p^2 - 8000p + 40000 = 0
factor this quadratic equationt to get:
p = 10.
raise the revenue to 42000 and do the samething you did when it was 40000 to get:
400p^2 - 8000p + 42000 = 0
factor this quadratic equation and you will find that the roots of this equation are complex and not real.
something is obviously wrong.
go back to the original revenue equation that was:
r = 8000p - 400p^2.
arrange the terms in descending order of degree to get:
r = -400p^2 + 8000p
set r equal to 0 and you get:
-400p^2 + 8000p = 0
the equation is now in standard form of ax^2 + bx + c = 0
a = -400
b = 8000
c = 0
the maximum / minimum value of a quadratic equation is determined by the formula of:
x = -b/2a
in this equation, p replaced x, so the maximum / minimum value of the quadratic equation is obtained by the formula of:
p = -b/2a.
solve for p to get:
p = -8000 / -800 = 10
the maximum value of the quadratic equation is when p = 10.
replace p in the revenue eqution of r = -400p^2 + 8000p to get:
r = 40,000.
that explains why the roots weren't real when you changed the revenue from 40,000 to 42,000.
you could not obtain a revenue of 42,000 because the maximum revenue of the equation was 40,000.
a graph of the equation also shows this to be true, as shown below.
y = 40,000 intersects the equation at x = 10.
y = 42,000 doesn't intersect the equation at all.
here's the graph.
