Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve optimisation problems (maximum/minimum values).
• Interpret and apply rates of change.
• Use first and second derivative tests.
• Solve related rates problems.
• Set up and solve applied optimisation problems.

Lesson summary

This week's focus is on consolidating your understanding of Calculus, specifically focusing on optimisation problems and rate of change, as these are frequently tested and often prove challenging for Grade 12 learners. These concepts are vital not only for the final examination but also for understanding various real-world phenomena. Calculus enables us to model and optimize various processes, from minimizing costs in manufacturing to maximizing profits in business. In South Africa, understanding optimization principles is crucial for efficient resource allocation and economic growth.

Lesson notes

2.1 Optimisation Problems Optimization problems involve finding the maximum or minimum value of a function subject to certain constraints. The general steps to solving these problems are: Understand the Problem: Read the problem carefully and identify what you are trying to maximize or minimize (the objective function). Identify any constraints given in the problem.

Draw a Diagram (if applicable): Visualizing the problem can often help in understanding the relationships between variables.

Introduce Variables: Assign variables to the quantities involved in the problem.

Formulate the Objective Function: Write the objective function in terms of the variables you've introduced. This function represents the quantity you want to maximize or minimize. Express the Objective Function in Terms of One Variable: Use the constraints to eliminate one or more variables from the objective function. This will leave you with a function of a single variable.

Find Critical Points: Differentiate the objective function with respect to the single variable and set the derivative equal to zero. Solve for the variable to find the critical points. Also, check for endpoints of the interval.

Test Critical Points: Use the first or second derivative test to determine whether each critical point corresponds to a maximum, a minimum, or neither.

Answer the Question: State the maximum or minimum value of the objective function, along with the value(s) of the variable(s) that produce this value.

Check for Reasonableness: Ensure that your solution makes sense in the context of the problem.

Example 1: A farmer in KwaZulu-Natal wants to fence off a rectangular garden next to a long stone wall. The garden must have an area of 100 m². What is the minimum amount of fencing required?

Solution: Understand the Problem: Minimize the amount of fencing (perimeter) given a fixed area.

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

Variables: Let x be the width of the garden and y be the length of the garden (perpendicular to the wall).

Objective Function: Minimize the fencing F = x + 2y (only three sides need fencing).

Constraint: Area A = x y = 100, so y = 100/x.

Express in One Variable: Substitute y in the fencing equation: F = x + 2(100/x) = x + 200/x.

Find Critical Points: dF/dx = 1 - 200/x² =

0. This gives x² = 200, so x = √200 = 10√2 (since x must be positive).

Test Critical Points: d²F/dx² = 400/x³. At x = 10√2, d²F/dx² > 0, so it's a minimum.

Answer: x = 10√2 meters. y = 100/x = 100/(10√2) = 5√2 meters. The minimum amount of fencing is F = 10√2 + 2(5√2) = 20√2 meters, or approximately 28.28 meters. 2.2 Rates of Change The derivative of a function f(x), denoted as f'(x) or df/dx, represents the instantaneous rate of change of f(x) with respect to x. In real-world applications, rate of change describes how one quantity changes with respect to another.

For example: Velocity: The rate of change of displacement (position) with respect to time.

Acceleration: The rate of change of velocity with respect to time.

Population Growth: The rate of change of population size with respect to time.

Related Rates: These problems involve finding the rate of change of one variable in terms of the rate of change of another variable, where both variables are related by some equation.

Example 2: A spherical balloon is being inflated at a rate of 100 cm³/s. How fast is the radius of the balloon increasing when the radius is 5 cm?

Solution: Understand the Problem: We're given dV/dt = 100 cm³/s and want to find dr/dt when r = 5 cm.

Equation: The volume of a sphere is V = (4/3)πr³.

Differentiate with respect to time: dV/dt = 4πr² (dr/dt).

Substitute: 100 = 4π(5)² (dr/dt).

Solve for dr/dt: dr/dt = 100 / (100π) = 1/π cm/s.

Example 3: Related Rates A ladder 5 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 2 m/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 3 m from the wall?

Solution: Understand the Problem: Given dx/dt = 2 m/s, find dy/dt when x = 3 m.

Equation: x² + y² = 5² (Pythagorean theorem)

Differentiate with respect to time: 2x(dx/dt) + 2y(dy/dt) = 0 Substitute: When x = 3, y = √(5² - 3²) = 4. 2(3)(2) + 2(4)(dy/dt) = 0 Solve for dy/dt: 12 + 8(dy/dt) = 0 => dy/dt = -12/8 = -3/2 m/s. 2.3 First and Second Derivative Tests First Derivative Test: This test helps identify local maxima and minima. If f'(x) changes from positive to negative at x = c, then f(x) has a local maximum at x = c. If f'(x) changes from negative to positive at x = c, then f(x) has a local minimum at x = c. If f'(x) does not change sign at x = c, then f(x) has neither a local maximum nor a local minimum at x = c.

Second Derivative Test: This test also helps identify local maxima and minima. If f'(c) = 0 and f''(c) > 0, then f(x) has a local minimum at x = c. If f'(c) = 0 and f''(c) 0, so it's a minimum.

Answer: x = 20 cm, h = 4000/20² = 10 cm.