Overview
This lecture explains how to solve an optimization problem to find the number of units that maximizes profit, given demand and cost functions.
Problem Setup
- Given a demand function (price as a function of units, x) and a cost function (cost as a function of x).
- Goal: Find the value of x (units) that yields maximum profit.
Understanding the Functions
- The demand function expresses price (P) as P = 3,000 - 2x.
- The cost function is C = 1,200x + 2,600.
- Higher x decreases price (demand function) and increases cost (cost function).
Step-by-Step Solution
- Step 1: Identify the equation to maximize; in this case, profit.
- Profit = Revenue - Cost.
- Step 2: Rewrite profit in terms of a single variable (x).
- Revenue = Price × Units Sold = (3,000 - 2x)x.
- Profit = (3,000 - 2x)x - (1,200x + 2,600).
- Expand and simplify: Profit = -2x² + 1,800x - 2,600.
- Step 3: Find critical values by taking the derivative and setting it to zero.
- d(Profit)/dx = -4x + 1,800; set to zero and solve for x: x = 450.
- Step 4: Use the second derivative test to confirm a maximum.
- Second derivative = -4 (constant negative), confirming a maximum at x = 450.
Conclusion / Final Answer
- The maximum profit occurs when 450 units are produced.
Key Terms & Definitions
- Optimization problem — Finding the value that maximizes or minimizes a particular function.
- Demand function — Relates price to quantity demanded (P = 3,000 - 2x).
- Cost function — Gives total production cost for x units (C = 1,200x + 2,600).
- Revenue — Total money received from sales; price times quantity sold.
- Profit — Difference between revenue and cost.
- Derivative — Measures the rate of change; used to find maxima/minima.
- Critical value — Input where the derivative equals zero; potential max/min point.
- Second derivative test — Used to determine if a critical value is a maximum or minimum.
Action Items / Next Steps
- Practice solving other optimization problems using different demand and cost functions.
- Review the steps for rewriting functions in terms of one variable.
- Apply the second derivative test to confirm maxima or minima in similar problems.