Montgomery County Community College - MAT 142: Cost-Revenue Analysis

This assignment provides a detailed analysis of cost, revenue, and profit functions. It begins by determining the minimum value of the average cost function using calculus. The relationship between marginal cost and average cost is explored graphically, noting the increasing nature of marginal cost versus the decreasing-then-increasing nature of average cost. Profit maximization is analyzed for three different revenue functions, revealing scenarios of negative, zero, and positive maximum profit. Graphical representation of marginal revenue functions alongside cost functions is used to interpret these profit outcomes. Finally, the concepts of under allocation, over allocation, and efficient allocation are discussed in relation to the graphical analysis, providing a comprehensive understanding of cost-revenue dynamics. Desklib offers a wide range of study resources, including past papers and solved assignments, to support students in mastering these concepts.

Q1)
The average cost function = C ( x )
x = 2 x2+50
x
= 2 x+ 50
x
Finding the minimum value by using differential approach
d ( C( x )
x )
dx =2− 50
x2 =0
x2−50=0
x2 = 50
x = 5− ¿+¿ ¿ ¿
this means that 5 must be an extreme point
d2
( C ( x)
x )
d x2 |x=5 = 100
x3 |x=5 = 0.8
The double differentiation is positive at extreme points, it will be minimum then
The minimum value of average cost function
C ( 5 )
5 =2∗5+ 50
5 =20
Q2)
Marginal cost function is determined by differentiating the cost function
MC(x) = d (2 x2 +50)
dx
= 4x
From question 1
C ( x )
x =2 x + 50
x
Plotting
x 0 5 10 15
MC(x) = 4x 0 20 40 60
x 0 5 10 15
C ( x )
x =2 x + 50
x
0 20 25 33.33

0 2 4 6 8 10 12 14 16
0
10
20
30
40
50
60
70
MC(x)
average cost
x
y
What is noticed is that the marginal function is always increasing in the domain while the
average cost function is first decreasing and then increasing
Q3)
1st revenue
R(x) = 10x
The function of profit will be given
P(x) = R(x) – C(x)
P(x) = 10x – 2x2 – 50
Finding the maximum profit in order to know if it is positive or negative
Using double differentiation
d (P ( x ) )
dx =10−4 x=0
x = 2.5
the critical point is 2.5
d (P ( x ) )
d x2 =−4
At the second differential is negative, the critical point is maximum

Pmax = P(x)|x=2.5 = 10*2.5 – 2*2.52 – 50 = -37.5
The maximum profit is negative
Using similar approach for revenue 2 and revenue 3
Revenue 2
R(x) = 20x
The function of profit will be given
P(x) = R(x) – C(x)
P(x) = 20x – 2x2 – 50
Finding the maximum profit in order to know if it is positive or negative
Using double differentiation
d (P ( x ) )
dx =20−4 x=0
x = 5
the critical point is 5
d (P ( x ) )
d x2 =−4
At the second differential is negative, the critical point is maximum
At the second differential is negative, the critical point is maximum
Pmax = P(x)|x=5 = 20*5 – 2*52 – 50 = 0
The maximum profit is zero
Revenue 3
R(x) = 30x
The function of profit will be given
P(x) = R(x) – C(x)
P(x) = 30x – 2x2 – 50
Finding the maximum profit in order to know if it is positive or negative
Using double differentiation
d ( P ( x ) )
dx =30−4 x=0
x = 5
the critical point is 7.5
d (P ( x ) )
d x2 =−4
At the second differential is negative, the critical point is maximum
At the second differential is negative, the critical point is maximum
Pmax = P(x)|x=7.5 = 30*7.5 – 2*7.52 – 50 = 62.5

Therefore,
Pmax = 62.5
The maximum profit is positive
Q4)
0 2 4 6 8 10 12 14 16
0
10
20
30
40
50
60
70
MC(x)
average coast
R(x) = 10
R(x) = 20
R(x)=30
x
y
Q5)
For first case y = 10 under allocation
For the second case y = 20 is efficient allocation
For the third case y = 30 is over allocation