Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply differentiation to find stationary points.
• Formulate functions for real-world scenarios.
• Solve optimization problems using calculus.
• Interpret results in the context of the problem.

Lesson summary

Differential calculus forms the bedrock of understanding change and rates of change. This week, we'll be focusing on applying differentiation rules to solve optimization problems. Optimization is the process of finding the maximum or minimum value of a function, often subject to constraints. Think about a farmer trying to maximize crop yield within the limitations of their land and resources, or a business aiming to minimize production costs while maintaining quality. These are real-world problems optimization can solve. In South Africa, optimization can be crucial for efficient resource allocation in industries like agriculture, manufacturing, and logistics.

Lesson notes

2.1 Stationary Points: A stationary point of a function f(x) is a point where the derivative, f'(x), is equal to zero. These points represent where the gradient of the function is horizontal.

Stationary points can be classified as: Local Maximum: A point where the function reaches a local maximum value. The derivative changes from positive to negative at this point. f''(x) 0 at the minimum point.

Point of Inflection: A point where the concavity of the function changes. The derivative might be zero at this point, but the second derivative is zero (or undefined) and changes sign around the point. f''(x) = 0 and changes sign at the point of inflection. The first derivative test will show a change in slope i.e. positive to negative for max and vice versa for minimum. 2.2 Optimization Problems: Optimization problems involve finding the maximum or minimum value of a function subject to certain constraints. The general steps to solve optimization problems are: Understand the problem: Read the problem carefully and identify the quantity to be optimized (maximized or minimized).

Draw a diagram (if applicable): A visual representation can often help clarify the relationships between variables.

Introduce variables: Define variables to represent the relevant quantities in the problem.

Formulate the function: Express the quantity to be optimized as a function of the variables. This is often the most challenging step.

Identify Constraints: Find any constraints (limitations) on the variables. These constraints will often be expressed as equations.

Reduce to a single variable: Use the constraints to eliminate one or more variables from the function to be optimized. This will result in a function of a single variable.

Find stationary points: Differentiate the function with respect to the single variable and set the derivative equal to zero to find the stationary points.

Determine the nature of stationary points: Use the second derivative test or the first derivative test to determine whether each stationary point is a maximum, a minimum, or a point of inflection.

Check endpoints (if applicable): If the domain of the function is restricted (e.g., by physical constraints), check the value of the function at the endpoints of the domain.

Answer the question: State the maximum or minimum value of the quantity being optimized and the values of the variables at which this occurs. 2.3 Worked

Examples: Example 1: A farmer in KwaZulu-Natal wants to fence off a rectangular vegetable garden next to a long wall. The farmer has 100 meters of fencing. What are the dimensions of the garden that maximize the area enclosed?

Solution: Understand the problem: Maximize the area of a rectangle with a fixed amount of fencing used for only three sides.

Diagram: (Imagine a rectangle with one long side being the wall).

Variables: Let l be the length of the garden parallel to the wall and w be the width of the garden perpendicular to the wall.

Function: The area to be maximized is A = l w.

Constraint: The perimeter of the three sides is l + 2w =

1

0

0. Reduce to a single variable: Solve the constraint for l: l = 100 - 2w.

Substitute this into the area equation: A = (100 - 2w)w = 100w - 2*w² Find stationary points: dA/dw = 100 - 4w =

0. Solving for w, we get w =

2

5. Determine the nature: d²A/dw² = -4, which is negative.

Therefore, w = 25 corresponds to a maximum.

Find l: l = 100 - 2(25) =

5

0. Answer: The dimensions of the garden that maximize the area are length = 50 meters and width = 25 meters. The maximum area is 50 * 25 = 1250 square meters.

Example 2: A soft drink company in Gauteng wants to manufacture cylindrical cans to hold a volume of 330ml (330 cm³). What dimensions (radius and height) should the can have to minimize the amount of material used (surface area)?

Solution: Understand the problem: Minimize the surface area of a cylinder with a fixed volume.

Diagram: (Imagine a cylinder).

Variables: Let r be the radius and h be the height of the can.

Function: The surface area to be minimized is A = 2πr² + 2πr*h.

Constraint: The volume is πr²h =

3

3

0. Reduce to a single variable: Solve the constraint for h: h = 330 / (πr²). Substitute this into the surface area equation: A = 2πr² + 2πr(330 / (πr²)) = 2πr² + 660/r.

Find stationary points: dA/dr = 4πr - 660/r² =

0. Multiplying by r², we get 4πr³ - 660 =

0. Solving for r, we get r³ = 660 / (4π) = 165 / π.

Therefore, r = (165/π)^(1/3) ≈ 3.76 cm.

Determine the nature: d²A/dr² = 4π + 1320/r³. Since r is positive, d²A/dr² is positive, indicating a minimum.

Find h: h = 330 / (π(3.76)²) ≈ 7.42 cm.

Answer: The dimensions that minimize the material used are radius ≈ 3.76 cm and height ≈ 7.42 cm.

Example 3: Find the point on the curve y = x² that is closest to the point (3, 0).

Solution: Understand the problem: Minimize the distance between a point on the parabola y=x² and the point (3,0).

Variables: Let (x, y) be a point on the curve y = x².