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
Answer:
$120,000
Explanation:
Total amount for inclusion in determining Mill Corp's net income or loss is as follows.
- Net loss from disposal of business segment = $100,000
- Property tax for 6 months to June 30, 20x5= $40,000 * 0.5 = $20,000
Therefore, total amount for inclusion = $100,000 + $20,000 = $120,000.
Answer: Labor
Explanation: Human Resources (labor) The quantity and quality of human effort directed toward producing goods and services (also called labor).
Answer
The answer and procedures of the exercise are attached in the following archives.
Explanation
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