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Answer Key: Grade 12 Calculus – Applications of Derivatives

Answer Key
Grade 12
English

Answer Key: Grade 12 Calculus – Applications of Derivatives

Answer Key: Grade 12 Calculus – Applications of Derivatives

This resource provides the solutions and grading criteria for the assessment on Applications of Derivatives: Optimization and Rate of Change.


Section 1: Multiple Choice & Short Answer

Question 1: Multiple Choice
  • Question: If $f(x) = x^3 - 3x^2 + 5$, at what value of $x$ does the local minimum occur?
  • Answer: C) $x = 2$
  • Explanation: The derivative is $f'(x) = 3x^2 - 6x$. Setting $f'(x) = 0$ gives $3x(x - 2) = 0$, so $x = 0$ and $x = 2$. Using the second derivative test, $f''(x) = 6x - 6$. Evaluating at critical points: $f''(0) = -6$ (local max) and $f''(2) = 6$. Since $f''(2) > 0$, there is a local minimum at $x = 2$.
Question 2: Fill in the Blank
  • Question: The derivative of a position function $s(t)$ represents the ________ of the object, while the second derivative represents the ________.
  • Answer: Velocity; Acceleration.
  • Grading Note: Both terms must be correct for full credit.
Question 3: True or False
  • Question: If $f'(c) = 0$, then there must be a local maximum or minimum at $x = c$.
  • Answer: False.
  • Correction: $x = c$ is a critical point, but it could be a point of inflection (e.g., $f(x) = x^3$ at $x = 0$ where the derivative is zero but there is no local extremum).


Section 2: Optimization Problem (Exemplar Solution)

Question 4: Open-Ended
  • Question: A rectangular enclosure is to be built next to a straight stone wall. 120 meters of fencing is available for the other three sides. Determine the dimensions that will maximize the area of the enclosure.

Model Answer (Exemplar)

Step 1: Define Variables and Constraints Let $L$ be the length of the side parallel to the wall and $W$ be the width (the two sides perpendicular to the wall). The total fencing used is given by the equation:

We want to maximize the Area ($A$):

Step 2: Express Area as a Single Variable Function Rearrange the constraint for $L$:

Substitute this into the Area formula:


Step 3: Find the Critical Points Take the derivative of $A$ with respect to $W$:

Set $A'(W) = 0$ to find the critical point:

Step 4: Verify Maximum and Solve for Length Using the second derivative test: $A''(W) = -4$. Since $A'' < 0$, the function is concave down, confirming $W = 30$ is a maximum point. Substitute $W = 30$ back to find $L$:

Final Answer: The dimensions that maximize the area are a width of 30 meters and a length of 60 meters, resulting in a maximum area of $1,800 \text^2$.


Grading Rubric & Evaluation Criteria

Criteria
Level 4 (Exemplary)
Level 3 (Proficient)
Level 1-2 (Developing)
Knowledge & Understanding
Accurately identifies the objective function and constraints with no errors.
Identifies objective function and constraints with minor errors.
Misidentifies variables or relationship between equations.
Application
Correctly applies differentiation rules and solves for critical values accurately.
Correctly applies differentiation but makes minor calculation errors.
Incorrect use of derivative or fails to find critical values.
Communication
Logic is clearly sequenced with units included and final statement provided.
Logic is mostly clear; may be missing units or concluding sentence.
Steps are disorganized or missing; final answer is unclear.
Teacher Guidance for Partial Credit:
  • Award 1 point for the correct constraint equation ($L + 2W = 120$).
  • Award 1 point for the correct Area function in one variable.
  • Award 2 points for the correct derivative and finding $W = 30$.
  • Award 1 point for verifying the maximum and providing the final dimensions.