Question
A stereo manufacturer determines that in order to sell x units of a new stereo, the price per unit, in rupees, must be p(x) = 1000-x. The manufacturer also determines that the total cost of producing x units is given by C(x) = 3000+20x.
a) Find the total revenue R(x).
b) Find the total profit P(x).
c) How many units must the manufacturer produce and sell in order to maximise profit?
d) What is the maximum profit?
e) What price per unit must be charged in order to make this maximum profit?
Answer :
Word Count : 400
Let's break down the problem step by step. ### Given: - Price per unit: \( p(x) = 1000 - x \) - Cost of production: \( C(x) = 3000 + 20x \) ### a) Find the total revenue \( R(x) \). Revenue is the product of the number of units sold and the price per unit. Therefore: \[ R(x) = x \cdot p(x) = x \cdot (1000 - x) \] Simplifying: \[ R(x) = 1000x - x^2 \] ### b) Find the total profit \( P(x) \). Profit is given by the difference between total revenue and total cost: \[ ____ _____ _________ ________ ____ ___.
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Let's break down the problem step by step. ### Given: - Price per unit: \( p(x) = 1000 - x \) - Cost of production: \( C(x) = 3000 + 20x \) ### a) Find the total revenue \( R(x) \). Revenue is the product of the number of units sold and the price per unit. Therefore: \[ R(x) = x \cdot p(x) = x \cdot (1000 - x) \] Simplifying: \[ R(x) = 1000x - x^2 \] ### b) Find the total profit \( P(x) \). Profit is given by the difference between total revenue and total cost: \[ ____ _____ _________ ________ ____ ___.
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