Question
A company manufactures two products, A and B. The
contribution per unit for A is ₹50 and for B is ₹60. Each unit of A requires 4 machine hours, and B requires 6 machine hours. Total machine hours available = 1,400. Fixed costs = ₹4,000. If the company can sell a maximum of 200 units of each, what is the maximum profit possible under machine hour constraints?Solution
Contribution per hour: A = 50/4 = ₹12.5, B = 60/6 = ₹10 → A has higher contribution per hour so first of all the machine hours will be utilized in manufacturing of product A. Max possible units = A: 200 units × 4 = 800 hrs Remaining = 1,400 – 800 = 600 hrs → B = 600/6 = 100 units Total contribution = (200×50) + (100 ×60) = ₹10,000 + ₹6,000 = ₹16,000 Profit = Contribution – Fixed = ₹16,000 – ₹4,000 = ₹12,000
cos (sec-1 x) is equal to –
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