Overview
This lecture explains how to solve an optimization problem by finding the number of units that results in maximum profit using given demand and cost functions.
Problem Setup
- Given two functions: the demand (price) function and the cost function.
- Goal: Find the value of x (units produced) that maximizes profit.
Understanding the Functions
- The demand function gives price P as a function of the number of units x.
- The cost function C is the cost to produce x units: C = 1,200x + 2,600.
- As x increases, price decreases (due to supply) and cost increases (due to production).
Steps for Optimization
1. Find the Equation to Maximize
- Profit = Revenue – Cost.
- Revenue is the total money received from selling x units.
2. Express Profit in One Variable
- Revenue = Price × Number of units = P × x.
- From the demand function: P = 3,000 – 2x.
- Substitute: Revenue = (3,000 – 2x)x.
- Substitute cost: C = 1,200x + 2,600.
- Profit = (3,000 – 2x)x – (1,200x + 2,600).
- Simplify profit: Profit = –2x² + 1,800x – 2,600.
3. Find Critical Values
- Take derivative: d(Profit)/dx = –4x + 1,800.
- Set derivative to zero: –4x + 1,800 = 0.
- Solve for x: x = 450.
4. Verify Maximum or Minimum
- Find second derivative: d²(Profit)/dx² = –4.
- Since –4 < 0, it’s a maximum.
5. State the Answer
- Maximum profit occurs at x = 450 units.
Key Terms & Definitions
- Demand function — Relates price to the quantity sold.
- Cost function — Gives the total cost for producing x units.
- Profit — The difference between revenue and cost.
- Revenue — Total money received from selling goods (price × quantity).
- Critical value — A value where the derivative is zero (potential maximum or minimum).
Action Items / Next Steps
- Review similar optimization problems for more practice.
- Practice taking derivatives and applying the second derivative test.