Answer: a) Let the revenue equation be R(x) = mx + b, where m and b are constants. Since the revenue equation is a polynomial of order 1, it can be written in this form.
Let the cost equation be C(x) = ax² + bx + c, where a, b, and c are constants. Since the cost equation is a polynomial of order 2, it can be written in this form.
We know that the graphs of the two equations meet at points A(1,1) and B(5,9). Therefore, we have the following two equations:
R(1) = m(1) + b = 1 (1)
R(5) = m(5) + b = 9 (2)
We can solve for m and b by solving the system of equations (1) and (2):
m = 2
b = -1
Therefore, the revenue equation is R(x) = 2x - 1.
We also know that the third point of the cost function is C(2,0), which gives us:
a(2)² + b(2) + c = 0
4a + 2b + c = 0
Using points A(1,1) and B(5,9), we can set up two more equations:
a(1)² + b(1) + c = 1 (3)
a(5)² + b(5) + c = 9 (4)
Simplifying (3) and (4), we get:
a + b + c = 1 (5)
25a + 5b + c = 9 (6)
Subtracting (5) from (6), we get:
24a + 4b = 8
6a + b = 2
Multiplying (5) by 6 and subtracting it from (6), we get:
19a + 3b = 3
Solving these two equations simultaneously, we get:
a = -1/5
b = 13/5
Substituting these values of a and b into equation (5), we get:
c = 3/5
Therefore, the cost equation is C(x) = (-1/5)x² + (13/5)x + (3/5).
b) The common solutions to the revenue and cost equations are the points where the company breaks even, i.e. where the revenue earned is equal to the cost incurred. These points are important because they represent the minimum amount of sales the company needs to make in order to avoid making a loss. In this case, the common solutions represent the points where the company earns just enough revenue to cover its costs.
Explanation: