Introduction
This question is based on concepts from business mathematics involving revenue, cost, and profit functions. We are given a pricing function and a cost function. We need to calculate the total revenue, total profit, and then determine the optimal number of units to maximize profit along with the corresponding maximum profit and the price per unit at that point.
Given Information
- Price per unit: p(x) = 1000 – x
- Cost function: C(x) = 3000 + 20x
- x: Number of units sold
a) Finding the Total Revenue R(x)
Total revenue R(x) is given by:
R(x) = Price × Quantity = p(x) × x = (1000 – x)x = 1000x – x²
b) Finding the Total Profit P(x)
Total profit is:
P(x) = Revenue – Cost
We substitute R(x) and C(x):
P(x) = (1000x – x²) – (3000 + 20x)
P(x) = 1000x – x² – 3000 – 20x = -x² + 980x – 3000
c) How Many Units to Maximize Profit?
The profit function is a quadratic function of the form:
P(x) = -x² + 980x – 3000
This is a downward opening parabola. The maximum value occurs at the vertex.
Vertex of a parabola ax² + bx + c is at x = -b/2a
Here, a = -1, b = 980
x = -980 / (2 × -1) = 490
Therefore, the manufacturer must produce and sell 490 units to maximize profit.
d) What is the Maximum Profit?
Substitute x = 490 into the profit function:
P(490) = -(490)² + 980×490 – 3000
P(490) = -240100 + 480200 – 3000 = 237100
Maximum profit = Rs. 2,37,100
e) What Price Per Unit for Maximum Profit?
We use the price function:
p(x) = 1000 – x
p(490) = 1000 – 490 = Rs. 510
So, the price per unit to make maximum profit is Rs. 510
Conclusion
This question illustrates how we use basic calculus and algebra to solve practical business problems. Here’s a summary of the results:
- Total Revenue: R(x) = 1000x – x²
- Total Profit: P(x) = -x² + 980x – 3000
- Units to Maximize Profit: 490 units
- Maximum Profit: Rs. 2,37,100
- Price per Unit: Rs. 510
By using mathematical modeling, companies can decide the ideal production quantity and pricing strategy to maximize their gains.