Derivative NEB 12 Business Mathematics

Question 1: Stationary Points and Inflection

Question: Find the stationary points and point of inflection of y = x^3 - 3x^2 - 9x + 15.

Step 1: Find the First Derivative (Stationary Points)
- First derivative: 3x^2 - 6x - 9.
- Set to zero: 3(x^2 - 2x - 3) = 0.
- Solving this quadratic gives x = 3 and x = -1.
- Substitute back to find y: Stationary points are (3, -12) and (-1, 20).

Step 2: Find the Second Derivative (Point of Inflection)
- Second derivative: 6x - 6.
- Set to zero: 6x = 6, so x = 1.
- Substitute back to find y: y = (1)^3 - 3(1)^2 - 9(1) + 15 = 4.
- Result: Point of Inflection is (1, 4).

Trick: Think of stationary points as "hills and valleys" and the inflection point as where the slope "stops turning" and starts going the other way.

Question 17: Profit Maximization

Question: Given demand P = 20 - Q and cost C = Q^2 + 8Q + 2, find output Q for max profit.

Step 1: Create the Profit Function
- Revenue (R) = Price * Quantity = (20 - Q) * Q = 20Q - Q^2.
- Profit = Revenue - Cost = (20Q - Q^2) - (Q^2 + 8Q + 2) = -2Q^2 + 12Q - 2.

Step 2: Differentiate Profit
- First derivative of Profit: -4Q + 12.
- Set to zero: -4Q = -12, so Q = 3.

Step 3: Verification
- Second derivative: -4.
- Since -4 is negative, profit is definitely Maximum at Q = 3.

Hint: Profit is highest when "Marginal Revenue" equals "Marginal Cost." You can solve it by setting the derivatives of R and C equal to each other!

Question 12: Area Maximization

Question: Find the maximum area of a rectangular plot that can be enclosed with a 120m rope.

Step 1: Use the Perimeter
- Perimeter = 2(Length + Width) = 120, so Length + Width = 60.
- Let Length = x, then Width = 60 - x.

Step 2: Create the Area Function
- Area (A) = x * (60 - x) = 60x - x^2.

Step 3: Solve
- Derivative dA/dx = 60 - 2x.
- Set to zero: 2x = 60, so x = 30.
- Max Area = 30 * 30 = 900 square meters.

Trick: When you have a fixed perimeter, a Square (where length equals width) always gives the maximum area!