Optimizing Costs and Profits for Desklib: A Study

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