Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve optimization problems (maximizing or minimizing).
• Determine and interpret rates of change.
• Apply calculus to model and solve real-world problems.
• Formulate mathematical models from word problems.

Lesson summary

This week, we delve deeper into differential calculus, specifically focusing on optimization problems and rates of change. Differential calculus provides us with powerful tools to find maximum and minimum values of functions. This has wide applications, from optimising profits in business to minimizing costs in engineering. Understanding rates of change allows us to model how quantities are changing with respect to time, which is vital in understanding trends in areas like population growth, economic development, and resource management. For South African learners, this topic is crucial for developing analytical and problem-solving skills applicable in various fields.

Lesson notes

Optimization Problems Optimization problems involve finding the maximum or minimum value of a function, often subject to certain constraints.

The key steps are: Identify the quantity to be optimized: What are you trying to maximize or minimize (e.g., profit, cost, area)? Define a variable for this quantity.

Identify the constraint(s): What limitations are you working with (e.g., available materials, fixed budget, area limitations)? Express these as equations or inequalities. Express the quantity to be optimized as a function of a single variable: Use the constraint(s) to eliminate variables until you have a function of only one variable.

Find the critical points: Differentiate the function with respect to the variable, set the derivative equal to zero, and solve for the variable. Also, consider endpoints of the domain, as the max/min could occur there. Determine whether the critical point(s) represent a maximum or minimum: Use the first or second derivative test.

Interpret the result: Answer the original question in the context of the problem.

Example 1: A farmer in KwaZulu-Natal wants to fence off a rectangular grazing area next to a long existing wall. He has 400 meters of fencing material. What are the dimensions of the rectangle that will maximize the area enclosed?

Quantity to be optimized: Area (A)

Constraint: Perimeter (P = 400 meters) Let the length of the rectangle (parallel to the wall) be x meters and the width be y meters. The area is given by A = x y. The perimeter is given by x + 2y = 400 (since one side is the wall and does not need fencing). From the perimeter equation, we can express x in terms of y: x = 400 - 2y Substitute this into the area equation: A = (400 - 2y)y = 400y - 2*y² Now, differentiate A with respect to y: dA/dy = 400 - 4y Set the derivative equal to zero: 400 - 4y = 0 => y = 100 Now, find x: x = 400 - 2(100) = 200 To confirm this is a maximum, we can use the second derivative test: d²A/dy² = -4, which is negative, indicating a maximum.

Therefore, the dimensions that maximize the area are x = 200 meters and y = 100 meters. The maximum area is A = 200 * 100 = 20000 square meters.

Example 2: A Johannesburg-based company wants to manufacture cylindrical metal tins to hold baked beans. Each tin must have a volume of 500 cm³. What dimensions (radius and height) should the company choose to minimize the amount of metal used (surface area)?

Quantity to be optimized: Surface Area (SA)

Constraint: Volume (V = 500 cm³) Let the radius of the tin be r cm and the height be h cm. The volume is given by V = πr²h = 500 The surface area is given by SA = 2πr² + 2πrh (two circles and the curved side) From the volume equation, we can express h in terms of r: h = 500/(πr²) Substitute this into the surface area equation: SA = 2πr² + 2πr(500/(πr²)) = 2πr² + 1000/r Now, differentiate SA with respect to r: dSA/dr = 4πr - 1000/r² Set the derivative equal to zero: 4πr - 1000/r² = 0 => 4πr³ = 1000 => r³ = 1000/(4π) => r = ∛(250/π) ≈ 4.30 cm Now, find h: h = 500/(π(4.30)²) ≈ 8.60 cm The second derivative test (d²SA/dr² = 4π + 2000/r³) will confirm that this is a minimum (positive value).

Therefore, the dimensions that minimize the surface area are approximately r = 4.30 cm and h = 8.60 cm. Rates of Change (Related Rates) Related rates problems involve finding the rate at which one quantity is changing in terms of the rates at which other quantities are changing.

The key steps are: Read the problem carefully: Understand the situation and what rates are given and what rate you need to find.

Draw a diagram: This can help visualize the relationships between the quantities.

Identify the variables and constants: Assign variables to the quantities that are changing and note any constants.

Write an equation relating the variables: This equation is often based on geometric formulas or physical principles. Differentiate the equation implicitly with respect to time (t): This will give you an equation relating the rates of change.

Substitute the given information: Plug in the known values for the variables and their rates of change.

Solve for the unknown rate: Solve the equation for the rate you are trying to find.

Interpret the result: Answer the original question in the context of the problem, including units.

Example 3: A water tank in the shape of an inverted cone is being filled with water at a rate of 0.2 m³/min. The cone has a height of 10 meters and a radius of 4 meters at the top. How fast is the water level rising when the water is 5 meters deep?

Variables: Volume (V), height (h), radius (r), time (t)

Constants: Cone's height (H = 10 m), Cone's radius (R = 4 m)

Given: dV/dt = 0.2 m³/min Find: dh/dt when h = 5 m The volume of a cone is V = (1/3)πr²h. We need to relate r and h.