If the selling price x is related to profit p by the equation p=5000x−125x^2
We can tell everything we need from the graph. So let's graph it:
p = 5000x − 125x²
Put it in descending order of powers of x:
p = −125x² + 5000x
We recognize this as a quadratic equation whose graph is a
parabola. Since the leading coefficient is negative it opens
downward.
The maximum point is the vertex.
We compare
p = −125x² + 5000x
to
y = ax² + bx + c
and determine that a = -125, b = 5000 and c = 0
The formula for the x-cordinate of the vertex is 

Substituting in p = -125x² + 5000x
p = -125(20)² + 5000(20)
p = -125(400) + 100000
p = -50000 + 100000
p = 50000
So the vertex is the point (20,50000)
We determine the x and p intercepts:
To determine the x-intercepts we set p = 0
0 = -125x² + 5000x
125x² - 5000x = 0
Divide through by 125
x² - 40x = 0
Factor out x
x(x - 40) = 0
Set both the factor x and the factor x - 40 equal to 0
x = 0; x - 40 = 0
x = 40
So the x-intercepts are (0,0) and (40,0)
To find the p-intercept, we set x = 0
p = -125(0)² + 5000(0)
p = 0
The p-intercept is (0,0), the origin.
i)For what range of values of x is the profit increasing?
on the interval 0 < x < 20
ii)For what range of values of x is the profit decreasing?
on the interval 20 < x < 40
iii)Determine the value of x that would yield maximum profit.
x = 20
iv)Determine the maximum profit.
p = 50000
Edwin