Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve optimisation problems (maximum/minimum values).
• Apply differentiation to analyze rates of change.
• Interpret the first and second derivatives in real-world scenarios.
• Model real-world situations using calculus.

Lesson summary

Differential calculus is a fundamental branch of mathematics that deals with rates of change and slopes of curves. In this week, we will delve deeper into optimisation problems and rates of change problems – crucial skills needed in various fields like engineering, economics, and even environmental science. Optimisation helps us find the best possible solutions, while understanding rates of change allows us to model and predict how things change over time. In South Africa, this knowledge can be applied to solve problems like optimising agricultural yields, designing efficient infrastructure projects (like roads or bridges), or managing resources sustainably.

Lesson notes

a)

Optimisation Problems: Optimisation problems involve finding the maximum or minimum value of a function, often subject to certain constraints. The general approach involves the following steps: Identify the Objective Function: This is the function you want to maximise or minimise (e.g., area, volume, profit, cost).

Identify the Constraints: These are limitations or conditions that must be satisfied (e.g., fixed perimeter, limited resources). Express the constraints as equations. Express the Objective Function in Terms of a Single Variable: Use the constraint equations to eliminate variables from the objective function. This will leave you with a function of one variable.

Find Critical Points: Differentiate the objective function with respect to the single variable and set the derivative equal to zero. Solve for the critical points.

Determine Maximum or Minimum: Use the first or second derivative test to determine whether each critical point corresponds to a maximum, minimum, or neither. Also, consider the endpoints of the domain if the domain is restricted.

Answer the Question: Make sure you answer the original question posed in the problem.

Example 1: A farmer in KwaZulu-Natal wants to fence off a rectangular field bordering a straight river. He has 100 meters of fencing. What are the dimensions of the field that will maximise the area enclosed?

Solution: Objective Function: Maximise the area, A = lw (length * width).

Constraint: The perimeter of the fencing is 100 meters. Since one side is along the river, we only need fencing for three sides: l + 2w =

1

0

0. Express A in terms of a single variable: From the constraint, l = 100 - 2w.

Substitute this into the area equation: A = (100 - 2w)w = 100w - 2w².

Find Critical Points: Differentiate A with respect to w: dA/dw = 100 - 4w. Set dA/dw = 0: 100 - 4w = 0 => w =

2

5. Determine Maximum: d²A/dw² = -

4. Since the second derivative is negative, the critical point w = 25 corresponds to a maximum.

Answer: When w = 25, l = 100 - 2(25) =

5

0. So, the dimensions that maximise the area are length 50 meters and width 25 meters. b)

Rates of Change: The derivative dy/dx represents the instantaneous rate of change of y with respect to x. It tells us how much y is changing at a particular value of x. In applied problems, these variables often represent physical quantities such as distance, time, velocity, and acceleration.

Velocity and Acceleration: If s(t) represents the position of an object at time t, then the velocity v(t) is the rate of change of position with respect to time: v(t) = ds/dt. The acceleration a(t) is the rate of change of velocity with respect to time: a(t) = dv/dt = d²s/dt².

Example 2: A stone is thrown vertically upwards from the top of a building with an initial velocity of 20 m/s. The height h(t) (in meters) of the stone above the ground after t seconds is given by h(t) = -5t² + 20t + 25. a) Find the velocity of the stone at time t. b) Find the time when the stone reaches its maximum height. c) Find the maximum height reached by the stone. d) Find the acceleration of the stone.

Solution: a)

Velocity: v(t) = dh/dt = -10t + 20 m/s. b)

Maximum Height: The stone reaches its maximum height when the velocity is zero: v(t) = 0 => -10t + 20 = 0 => t = 2 seconds. c)

Maximum Height: Substitute t = 2 into the height equation: h(2) = -5(2)² + 20(2) + 25 = -20 + 40 + 25 = 45 meters. d)

Acceleration: a(t) = dv/dt = -10 m/s². (Constant acceleration due to gravity). c)

Interpreting the Second Derivative: The second derivative, d²y/dx², tells us about the concavity of the function. If d²y/dx² > 0, the function is concave up (like a smile). If d²y/dx² 0 when 6x - 12 > 0 => x >

2. The function is concave up for x > 2. f''(x) x x =

2. The inflection point is at x =

2. The y-coordinate is f(2) = 2³ - 6(2²) + 5(2) = 8 - 24 + 10 = -

6. So the inflection point is (2, -6). Guided Practice (With Solutions)

Question 1: A rectangular garden is to be fenced off. The fencing for the three sides costs R50 per meter, and the fencing for the fourth side costs R100 per meter. If the area of the garden is to be 1000 m², find the dimensions of the garden that will minimise the cost of the fencing.

Solution: Let x and y be the dimensions of the garden. Let the side with cost R100 per meter have length x.

Cost: C = 100x + 50x + 50y + 50y = 150x + 100y. We want to minimise

C. Area: xy = 1000 => y = 1000/x.

Substitute y into the cost equation: C = 150x + 100(1000/x) = 150x + 100000/x.

Differentiate C with respect to x: dC/dx = 150 - 100000/x². Set dC/dx = 0: 150 - 100000/x² = 0 => 150x² = 100000 => x² = 100000/150 = 2000/3 => x = √(2000/3) ≈ 25.

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2. Then y = 1000/x = 1000/√(2000/3) = √(3000000/2000) = √1500 ≈ 38.

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3. The dimensions are approximately 25.82 m and 38.73 m. This minimizes the cost because the second derivative d²C/dx² = 200000/x³ which is positive, meaning that we have a minimum.

Question 2: A company in Gauteng produces solar panels.