Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve optimization problems by maximizing or minimizing functions.
• Apply optimization to geometric shapes.
• Solve rate of change (related rates) problems.
• Interpret solutions in the context of the problem.
• Formulate mathematical models from word problems.

Lesson summary

Differential calculus forms the bedrock of understanding change and rates of change, concepts vitally important in a multitude of real-world scenarios. In South Africa, understanding how things change is crucial for things like modelling population growth, optimizing agricultural yields given fluctuating rainfall, predicting financial market trends, and designing efficient infrastructure. This week's focus is on optimization problems and rate of change problems, building upon the differentiation techniques learned previously. We will apply our knowledge of finding derivatives to maximize or minimize certain quantities, like profit, cost, or area, and solve problems involving related rates.

Lesson notes

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

The general procedure involves: Understanding the problem: Read the problem carefully and identify the quantity to be optimized (maximized or minimized) and any constraints.

Formulating the mathematical model: Express the quantity to be optimized as a function of one or more variables. Use the constraints to eliminate variables and express the function in terms of a single variable.

Finding the critical points: Differentiate the function and find the critical points by setting the derivative equal to zero and solving for the variable. Also, check for points where the derivative is undefined.

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

Answering the question: State the maximum or minimum value and the corresponding value of the variable(s). Make sure to include units and interpret the solution in the context of the problem.

Example 1: Maximizing Area A farmer in KwaZulu-Natal has 200 meters of fencing to enclose a rectangular field. What dimensions will maximize the area of the field?

Solution: Understanding the problem: We want to maximize the area of a rectangle given a fixed perimeter.

Formulating the mathematical model: Let the length of the rectangle be `l` and the width be `w`. Area, `A = l w` (This is what we want to maximize) Perimeter, `P = 2l + 2w = 200` (This is the constraint) We can solve the perimeter equation for one variable, say `l`: `l = 100 - w` Substitute this into the area equation: `A = (100 - w) * w = 100w - w^2` Finding the critical points: Differentiate `A` with respect to `w`: `dA/dw = 100 - 2w` Set `dA/dw = 0`: `100 - 2w = 0` `2w = 100` `w = 50` Determining the maximum: Find the second derivative: `d^2A/dw^2 = -2` Since the second derivative is negative, we have a maximum at `w = 50`.

Answering the question: When `w = 50`, `l = 100 - 50 = 50` The dimensions that maximize the area are length = 50 meters and width = 50 meters. The maximum area is `A = 50 50 = 2500` square meters. Thus, a square field maximizes the area.

Example 2: Minimizing Cost A company in Gauteng wants to build a rectangular storage container with a volume of 100 cubic meters. The material for the base costs R80 per square meter, the material for the top costs R50 per square meter, and the material for the sides costs R60 per square meter. Find the dimensions of the container that will minimize the cost.

Solution: Understanding the problem: Minimize cost of a rectangular container given a fixed volume and different material costs.

Formulating the mathematical model: Let length = l, width = w, height = h. Volume, V = lwh = 100 (constraint) Cost, C = (80 lw) + (50 lw) + (60 2lh) + (60 * 2wh) = 130lw + 120lh + 120wh (minimize)

Solve for h in volume: h = 100/(lw)

Substitute into the cost equation: C = 130lw + 120l[100/(lw)] + 120w[100/(lw)] C = 130lw + 12000/w + 12000/l Finding the critical points: We need to use partial derivatives here, but since single variable calculus is focus, we will assume l = w to simplify. Then h = 100/l^2 Cost becomes: C = 130l^2 + 24000/l Differentiate: dC/dl = 260l - 24000/l^2 Set derivative to 0: 260l = 24000/l^2 => 260l^3 = 24000 => l^3 = 24000/260 => l = (2400/26)^(1/3) approximately 4.52 meters Determining minimum: Find the second derivative: d^2C/dl^2 = 260 + 48000/l^

3. This is positive so we have a minimum.

Answering the question: l = w = 4.52 meters, h = 100/(4.52)^2 = 4.89 meters The dimensions that minimize cost are approximately l = 4.52 m, w = 4.52 m, h = 4.89 m Rate of Change Problems (Related Rates) Rate of change problems involve finding the relationship between the rates of change of two or more variables that are related to each other.

The general procedure involves: Understanding the problem: Draw a diagram if possible and identify the variables whose rates of change are given and the variable whose rate of change you want to find.

Finding the relationship: Find an equation that relates the variables. This often involves geometric relationships, such as the Pythagorean theorem or trigonometric ratios.

Differentiating with respect to time: Differentiate both sides of the equation with respect to time `t`, using the chain rule. Remember that the derivative of a variable with respect to time represents its rate of change.

Substituting known values: Substitute the given values of the variables and their rates of change into the differentiated equation.

Solving for the unknown rate of change: Solve the equation for the rate of change you want to find.

Answering the question: State the value of the unknown rate of change, including units.

Example 3: Expanding Circle A circular oil spill is expanding. The radius of the spill is increasing at a rate of 2 meters per minute.