Mat 142 College Mathematics For Assessment Answers
Questions:
Answer:
The average cost function
Finding the minimum value by using differential approach
x2 = 50
x =
this means that 5 must be an extreme point
x=5 = x=5 = 0.8
The double differentiation is positive at extreme points, it will be minimum then
The minimum value of average cost function
Marginal cost function is determined by differentiating the cost function
MC(x) =
= 4x
From question 1
Plotting
|
x |
0 |
5 |
10 |
15 |
|
MC(x) = 4x |
0 |
20 |
40 |
60 |
|
x |
0 |
5 |
10 |
15 |
|
|
0 |
20 |
25 |
33.33 |
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
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
x = 2.5
the critical point is 2.5
At the second differential is negative, the critical point is maximum
Pmax = 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
x = 5
the critical point is 5
At the second differential is negative, the critical point is maximum
At the second differential is negative, the critical point is maximum
Pmax = 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
x = 5
the critical point is 7.5
At the second differential is negative, the critical point is maximum
At the second differential is negative, the critical point is maximum
Pmax = x=7.5 = 30*7.5 – 2*7.52 – 50 = 62.5
Therefore,
Pmax = 62.5
The maximum profit is positive
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
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