Machine A can produce 50 components a day while machine B only 40. The monthly maintenance cost for machine A is $1500 while that for machine B is $550. If each component generates an income of $10 what is the least number of days per month that the plant has to work to justify the usage of machine A instead of machine B?

(A) 6
(B) 7
(C) 9
(D) 10
(E) 11





M08-10

Machine A can produce 50 components a day while Machine B only 40. The monthly maintenance cost for Machine A is $1500 while the cost for Machine B is $550. If each component generates an income of $10, what is the least number of days per month that the plant has to operate to justify the usage of Machine A instead of Machine B?

A. 6
B. 7
C. 9
D. 10
E. 11


Monthly maintenance costs for machine A and for machine B are fixed. Meaning that even if the plant doesn't operate at all it'll still have these maintenance costs. The questions basically asks about minimum # of days (d) that plant should operate so that the profit from A is more than or equal to the profit from B.

Profit from machine A in d days: \(50*10*d-1,500\);

Profit from machine B in d days: \(40*10*d-550\);

\(50*10*d-1,500 \ge 40*10*d-550\), which leads to \(d \ge 9.5\). Hence the minimum # of days is 10.


Answer: D.

M08-10
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We are given that Machine A can produce 50 components per day, and that each component brings revenue of $10 per component. Thus, we know that Machine A’s daily revenue is (50)(10) = 500. We are also given that Machine A’s maintenance fee is $1500 per month.

Machine B produces 40 components per day and each component brings revenue of $10 per component. Thus, we know that Machine B’s daily revenue is (40)(10) = 400. Machine B’s maintenance fee is $550 dollars per month.

Since we need to justify the usage of Machine A, we need to determine the minimum number of days it will take until Machine A’s profit is greater than Machine B’s profit. We can let t = the number of days until this happens and create the following inequality:

50(10)(t) - 1500 > 40(10)(t) - 550

500t - 1500 > 400t - 550

100t > 950

t > 9.5

Thus, the minimum number of days needed is 10.

Answer: D
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