Optimizing Costs and Profits for Desklib: A Study

This study focuses on optimizing costs and profits for Desklib, an online library for study material. It covers topics such as the average cost function, marginal cost function, revenue, and profit to help make informed decisions. The study also includes a detailed analysis of the different revenue functions and their corresponding profit maximization points.

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