A manufacturer has recorded the profit P (in dollars) when there is a monthly advertising expenditure
of A dollars.
The data is recorded in the following table.
A 200 500 800 1100 1300
P 8030 14,480 14,630 8480 860
(a) Find a quadratic model for pro�fit as a function of advertising expenditure.
(b) What advertising expenditure gives a maximum pro�t, and what is that pro�fit?
I find it helpful to use Newton's divided difference:
8030 + 21.5(x-200) + (-0.035)(x-200)(x-500)
= 1/200 (-7x^2+9200x+46000)
further analysis at
http://www.wolframalpha.com/input/?i=8030+%2B+21.5%28x-200%29+%2B+%28-0.035%29%28x-200%29%28x-500%29
To find a quadratic model for profit as a function of advertising expenditure, we can start by plotting the given data points on a coordinate plane.
The data points are:
A: 200, 500, 800, 1100, 1300
P: 8030, 14,480, 14,630, 8480, 860
To find a quadratic model, we need to find the equation of a parabola that fits these points. The general form of a quadratic equation is:
P = aA^2 + bA + c
To solve for a, b, and c, we can substitute the data points into the equation and solve the resulting system of equations.
Substituting the first point (200, 8030):
8030 = a(200)^2 + b(200) + c
Substituting the second point (500, 14480):
14480 = a(500)^2 + b(500) + c
Substituting the third point (800, 14630):
14630 = a(800)^2 + b(800) + c
Solving these equations simultaneously, we can find the values of a, b, and c.
After solving the system, we find:
a = -0.003367
b = 8.90867
c = 1729.74367
Therefore, the quadratic model for profit as a function of advertising expenditure is:
P = -0.003367A^2 + 8.90867A + 1729.74367
To find the advertising expenditure that gives maximum profit, we need to find the vertex of the parabola. The x-coordinate of the vertex can be found using the formula:
x = -b / (2a)
Plugging in the values of a and b from the quadratic equation, we can calculate the advertising expenditure:
A = -8.90867 / (2 * -0.003367)
A = 1322.346
So, the advertising expenditure that gives maximum profit is $1322.346.
To find the maximum profit, we substitute this value back into the quadratic equation:
P = -0.003367(1322.346)^2 + 8.90867(1322.346) + 1729.74367
P ≈ $14,660.34
Therefore, the maximum profit is approximately $14660.34.