Answer:
Value of x maximising profit : x = 5
Explanation:
Cost : C(x) = x^3 - 6x^2 + 13x + 15 ; Revenue: R(x) = 28x
Profit : Revenue - Cost = R(x) - C(x)
28x - [x^3 - 6x^2 + 13x + 15] = 28x - x^3 + 6x^2 - 13x - 15
= - x^3 + 6x^2 + 15x - 15
To find value of 'x' that maximises total profit , we differentiate total profit function with respect to x & find that x value.
dTP/dx = - 3x^2 + 12x + 15 = 0 ► 3x^2 - 12x - 15 = 0
3x^2 + 3x - 15x - 15 = 0 ► 3x (x +1) - 15 (x + 1) = 0 ► (x+1) (3x-15) = 0
x + 1 = 0 ∴ x = -1 [Rejected, production quantity cant be negative] ;
3x - 15 = 0 ∴ 3x = 15 ∴ x = 15/3 = 5
Double derivate : d^2TP/dx^2 = - 6x + 12
d^2TP/dx^2 i.e - 6x + 12 at x = 5 is -6(5) + 12 = - 30+ 12 = -8 which is negative. So profit function is maximum at x = 5
A. The percentage of the labor force that is unemployed
*matches pairs to respective categories*
Answer:
Total cost= $60,800
Explanation:
Giving the following information:
For 4,000 units:
Unitary variable cost= $8
Unitary fixed cost= $5
<u>First, we need to calculate the total fixed cost:</u>
Total fixed cost= 5*4,000= $20,000
<u>Now, we can determine the total cost for 5,100 units:</u>
Total cost= 5,100*8 + 20,000
Total cost= $60,800
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