Question 2-9: Choosing the Most Economical Machine for Production Two curr...

(a) Rule 1 applies in this situation because total daily revenues (selling price per part times the number of parts sold per day) and total daily costs will vary depending on the machine chosen. Therefore, we should select the machine that will maximize the profit per day:

Profit per day = Revenue per day − Cost per day

= (Production rate)(Production hours)($12/part)
× [1 − (%rejected/100)]
− (Production rate)(Production hours)($6/part)
− (Production hours)($15/hour + $5/hour).

Machine A: Profit per day =\left(\frac{100 \text { parts }}{\text { hour }}\right)\left(\frac{7 \text { hours }}{\text { day }}\right)\left(\frac{\$ 12}{\text { part }}\right)(1-0.03) \\-\left(\frac{100 \text { parts }}{\text { hour }}\right)\left(\frac{7 \text { hours }}{\text { day }}\right)\left(\frac{\$ 6}{\text { part }}\right) \\-\left(\frac{7 \text { hours }}{\text { day }}\right)\left(\frac{\$ 15}{\text { hour }}+\frac{\$ 5}{\text { hour }}\right)

= $3,808 per day.

Machine B: Profit per day =\left(\frac{130 \text { parts }}{\text { hour }}\right)\left(\frac{6 \text { hours }}{\text { day }}\right)\left(\frac{\$ 12}{\text { part }}\right)(1-0.10) \\-\left(\frac{130 \text { parts }}{\text { hour }}\right)\left(\frac{6 \text { hours }}{\text { day }}\right)\left(\frac{\$ 6}{\text { part }}\right) \\-\left(\frac{6 \text { hours }}{\text { day }}\right)\left(\frac{\$ 15}{\text { hour }}+\frac{\$ 5}{\text { hour }}\right)

= $3,624 per day.

Therefore, select Machine A to maximize profit per day.

(b) To find the breakeven percent of parts rejected, X, for Machine B, set the profit per day of Machine A equal to the profit per day of Machine B, and solve for X:

Thus, X = 0.08, so the percent of parts rejected for Machine B can be no higher than 8% for it to be as profitable as Machine A.