Your "gross monthly income" is the amount you make BEFORE they take out any deductions.
Your "gross monthly income" is the amount you make AFTER they take out any deductions.
Answer:
a. 8,000 + 1,000 + 3.2Q
b. 27,000 + 3.2Q
c. 15,000 Units
Explanation:
a. The accounting cost function is shown below:-
Accounting cost function = Fixed Leasing and insurance cost + material cost and supplied cost
= 8,000 + 1,000 + 3.2Q
b. The economic cost function is shown below:-
Economic cost function = Accounting cost + Opportunity cost
= 9,000 + 3.2Q + 3*6,000
=27,000 + 3.2Q
c. The computation of break even point is shown below:-
Break even Point = Total Fixed Cost ÷ Price - Average Variable cost
= 27,000 ÷ 5 - 3.2
= 15,000 Units
Answer:
$3.20 per unit
Explanation:
In this question, we have to compare the cost between two cases
In the first case, the total cost per unit would be
= Direct materials per unit + direct labor per unit + overhead cost per unit
= $11 + $25 + $17
= $53
In the first case, the total cost per unit would be
= Purchase price + overhead cost
= $48.55 + $17 × 45%
= $48.55 + $7.65
= $56.20
So, the difference would be
= $56.20 - $53
= $3.20 per unit
Goodwill would be unable to measure the standardize tests.
Answer:
They should operate Mine 1 for 1 hour and Mine 2 for 3 hours to meet the contractual obligations and minimize cost.
Explanation:
The formulation of the linear programming is:
Objective function:

Restrictions:
- High-grade ore: 
- Medium-grade ore: 
- Low-grade ore: 
- No negative hours: 
We start graphing the restrictions in a M1-M2 plane.
In the figure attached, we have the feasible region, where all the restrictions are validated, and the four points of intersection of 2 restrictions.
In one of this four points lies the minimum cost.
Graphically, we can graph the cost function over this feasible region, with different cost levels. When the line cost intersects one of the four points with the lowest level of cost, this is the optimum combination.
(NOTE: it is best to start with a low guessing of the cost and going up until it reaches one point in the feasible region).
The solution is for the point (M1=1, M2=3), with a cost of C=$680.
The cost function graph is attached.