S/unit.The market survey indicates that teonapachizeia ith a capacity of 10 have

Question

S/unit.The market survey indicates that teonapachizeia ith a capacity of 10 have a fixed cost of: uits paying one demand toy which will sell for 10 Mum i not exce 0.00 to produce this tove eacj Machine AwoulGs 2 e of two 0 KSpa., and yield a pronr Capacity, Machine B would nave fixed costs o pr and yield a profit ot machin KSpa. r 30 k pDa. iug onee deman loy. which bade: Sed he fit of 30 s of 16 KS pa.. (Spa. Utilized at maximum KS pa Requied: Using the contribution maran an aon a) The breakeven sales amount for ely b) The sales level where either machine is c) The range ch machincalculate qually p achinertabitha than the other of sales where one machine is more profitable

Explanation / Answer

a).

Consider the given problem here there 2 machines to produce this toy each with the capacity of “10,000 units” pa, now the company has a choice of buying one of these 2 machines to produce the toy.

Now, for the 1st machine the “F1 = Fixed cost = $30K” pa, “1 = Profit = $30K” pa. Let’s assume that the “per unit variable cost” is “v1”.

So, at the maximum capacity, 1 = P*X1 – F1 – v1*X1, => $30K = $10*10,000 - $30K – v1*X1.

=> $60K = $10*10,000 – v1*X1, => v1*X1 = 10*10,000 – $60K = $100,000 – $60,000 = $40,000.

=> v1 = $40,000 / X1 = $40,000 / 10,000 = $4, => v1 = $4.

Now, for the 2nd machine the “F1 = Fixed cost = $16K” pa, “2 = Profit = $24K” pa. Let’s assume that the “per unit variable cost” is “v2”.

So, at the maximum capacity, 2 = P*X2 – F2 – v2*X2, => $24K = $10*10,000 - $16K – v2*10,000.

=> $24K = $10*10,000 - $16K – v2*10,000, => $24K + $16K = 100,000 – v2*10,000.

=> v2*10,000 = $100,000 – $40K = $60,000, => v2 = 60,000/10,000 = $6, => v2 = $6.

According to the “CMA”, the break-even sales for machine A is, “X1 = F1 / (P – v1) = $30K/(10 - 4) = 30,000/6=5,000.

Similarly, for machine B the break-even sales is, “X2 = F2 / (P – v2) = $16K / (10 - 6) = 16,000 / 4 = 4,000.

So, for the "Machine A" the break-even sales is "5,000" and for "Machine B" the sales is "4,000".

b).

Let’s assume that if “X1=X2=X” both the machines are equally profitable, => at this level of sales the profit will be same for both machines.

=> P*X – F1 – v1*X = P*X – F2 – v2*X, => F1 + v1*X = F2 + v2*X, => $30K + $4*X = $16K + $6*X,

=> $30K - $16K = $6*X - $4*X = $2*X, => X = $14,000 / $2 = 7,000, => X = 7,000.

So, the level of sales where both machines are equally profitable is “X1 = X2 = X = 7,000”.

c).

Now, here we have to find out the range of sales where “Machine A” is more profitable compared to “Machine B”.

=> 1 > 2, => P*X – F1 – v1*X > P*X – F2 – v2*X, => F1 + v1*X < F2 + v2*X, => $30K + $4*X < $16K + $6*X,

=> $30K - $16K < $6*X - $4*X = $2*X, => X > $14K / $2 > 7,000, => X > 7,000.

=> If the level of sales is more than “7,000”, then “Machine A” is more profitable compared to “Machine B”. Similarly if “X < 7000”, => “Machine B” is more profitable compared to “Machine A”.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.