Grade 11
  1. Algebra  1.1. Sequences and Series  1.1.1. Arithmetic Sequence  1.1.2. Arithmetic Series  1.1.3. Geometric Sequence  1.1.4. Geometric Series  1.1.5. Convergence and Divergence  1.1.6. Sum to Infinity  1.1.7. Sigma Notation  1.2. Complex Numbers  1.2.1. Imaginary and Complex Numbers  1.2.2. Operations with Complex Numbers  1.2.3. Polar and Cartesian Forms  1.2.4. Complex Conjugates  1.2.5. Modulus and Argument  1.2.6. De Moivre's Theorem  1.2.7. Powers and Roots of Complex Numbers  1.3. Binomial Theorem  1.3.1. Expansion of Binomials  1.3.2. General Term  1.3.3. Binomial Coefficients  1.3.4. Applications of Binomial Theorem  1.3.5. Pascal's Triangle
  2. Functions and Graphs  2.1. Types of Functions  2.1.1. Polynomial Functions  2.1.2. Rational Functions  2.1.3. Exponential Functions  2.1.4. Logarithmic Functions  2.1.5. Trigonometric Functions  2.1.6. Piecewise Functions  2.2. Transformations of Functions  2.2.1. Translations  2.2.2. Reflections  2.2.3. Dilations and Stretches  2.2.4. Rotations  2.3. Inverse Functions  2.3.1. Finding Inverses  2.3.2. Graphs of Inverse Functions  2.3.3. Properties of Inverse Functions  2.3.4. Restrictions for Inverses
  3. Trigonometry  3.1. Trigonometric Ratios and Identities  3.1.1. Basic Trigonometric Ratios  3.1.2. Pythagorean Identities  3.1.3. Sum and Difference Formulas  3.1.4. Double Angle Formulas  3.1.5. Half Angle Formulas  3.1.6. Product-to-Sum and Sum-to-Product Formulas  3.2. Trigonometric Equations  3.2.1. Solving Trigonometric Equations  3.2.2. General Solutions in Trigonometric Equations  3.3. Graphs of Trigonometric Functions  3.3.1. Sine Graph  3.3.2. Cosine Graph  3.3.3. Tangent Graph  3.3.4. Amplitude and Period Adjustments  3.3.5. Phase Shifts in Graphs of Trigonometric Functions  3.4. Applications of Trigonometry  3.4.1. Laws of Sines and Cosines  3.4.2. Area of Triangles  3.4.3. Solving Triangles  3.4.4. Bearings and Navigation
  4. Calculus  4.1. Limits and Continuity  4.1.1. Concept of a Limit  4.1.2. One-Sided Limits  4.1.3. Limits at Infinity  4.1.4. Continuity of a Function  4.1.5. Types of Discontinuities  4.2. Differentiation  4.2.1. Definition of Derivative  4.2.2. Rules of Differentiation  4.2.3. Higher Order Derivatives  4.2.4. Chain Rule  4.2.5. Product Rule  4.2.6. Quotient Rule  4.3. Applications of Differentiation  4.3.1. Rate of Change  4.3.2. Tangents and Normals  4.3.3. Maxima and Minima  4.3.4. Increasing and Decreasing Functions  4.3.5. Optimization Problems  4.3.6. Curve Sketching  4.3.7. Related Rates  4.4. Integration  4.4.1. Antiderivatives  4.4.2. Indefinite Integrals  4.4.3. Definite Integrals  4.4.4. Fundamental Theorem of Calculus  4.4.5. Techniques of Integration  4.4.6. Area Under a Curve  4.5. Applications of Integration  4.5.1. Area Between Curves  4.5.2. Volumes of Solids of Revolution  4.5.3. Average Value of a Function
  5. Vectors and Matrices  5.1. Vectors  5.1.1. Representation of Vectors  5.1.2. Vector Operations  5.1.3. Dot Product  5.1.4. Cross Product  5.1.5. Applications in Geometry  5.1.6. Projection of Vectors  5.2. Matrices  5.2.1. Types of Matrices  5.2.2. Matrix Operations  5.2.3. Determinants  5.2.4. Inverse of a Matrix  5.2.5. Solving Systems of Linear Equations  5.2.6. Cramer's Rule
  6. Probability and Statistics  6.1. Probability  6.1.1. Basic Probability Concepts  6.1.2. Conditional Probability  6.1.3. Independent and Dependent Events  6.1.4. Bayes' Theorem  6.1.5. Combinatorial Probability  6.2. Random Variables  6.2.1. Discrete Random Variables  6.2.2. Continuous Random Variables  6.2.3. Probability Distributions  6.3. Probability Distributions  6.3.1. Binomial Distribution  6.3.2. Normal Distribution  6.3.3. Poisson Distribution  6.3.4. Uniform Distribution  6.4. Statistics  6.4.1. Measures of Central Tendency  6.4.2. Measures of Dispersion  6.4.3. Data Collection and Representation  6.4.4. Correlation and Regression  6.4.5. Sampling Techniques
  7. Coordinate Geometry  7.1. Straight Lines  7.1.1. Slope and Intercept Form  7.1.2. Distance Formula  7.1.3. Midpoint Formula  7.1.4. Angle Between Lines  7.1.5. Equation of Lines  7.2. Conic Sections  7.2.1. Circle  7.2.2. Parabola  7.2.3. Ellipse  7.2.4. Hyperbola  7.2.5. Properties of Conics  7.2.6. Parametric Equations of Conics  7.2.7. Polar Coordinates of Conics
  8. Mathematical Reasoning  8.1. Logic  8.1.1. Statements and Logical Connectives  8.1.2. Truth Tables  8.1.3. Logical Equivalence  8.1.4. Quantifiers  8.1.5. Proof Techniques  8.2. Proofs  8.2.1. Direct Proof  8.2.2. Indirect Proof  8.2.3. Proof by Contradiction  8.2.4. Proof by Mathematical Induction

Grade 11 → Calculus → Applications of Differentiation ↓


Optimization Problems


Optimization problems are an important part of calculus, especially when dealing with applications of differentiation. These problems involve finding the best solution, such as maximizing profits, minimizing costs, or utilizing resources in the best possible way. In mathematics, optimization focuses on finding the maximum or minimum value of a function within a given set of constraints.

Understanding adaptation through differentiation

Differentiation helps us understand how functions change. By finding the derivative of a function, we can identify points where the function has maximum or minimum values. These points are called extremum points and are essential in optimization problems.

Let f(x) be a function. The derivative of f(x) is denoted as f'(x).

When f'(x) = 0, this indicates possible maximum or minimum points, called critical points.

Using derivatives to solve optimization problems

The process of solving optimization problems usually includes the following steps:

  1. Identify the problem: Clearly define what you are trying to optimize. Determine if you need to maximize or minimize something, such as area, volume, profit, or cost.
  2. Set up a function: Write a mathematical function for the quantity you want to optimize. This function is often called the objective function.
  3. Determine the constraints. Constraints are conditions that must be met. These could be size limitations, budget constraints, etc. You will set up equations or inequalities to represent these constraints.
  4. Find the derivative: Differentiate the objective function with respect to the variables involved.
  5. Solve for critical points: Set the derivative equal to zero and solve for the variable. This step helps find potential maximum or minimum points.
  6. Perform interval analysis or use the second derivative test: Determine the nature of each critical point using interval analysis or the second derivative test.
  7. Conclusion: Draw conclusions based on your analysis and answer the original problem.

Example: Optimizing the area of a rectangle

Consider optimizing the area of a rectangle given a certain perimeter. Let's say the perimeter is 20 units.

Objective Function: We have to maximize the area of the rectangle.

If l is the length and w is the width of the rectangle, then Perimeter, P = 2l + 2w = 20.

From the circumference equation:

w = 10 - l

Area, A = l * w.

Substitute w = 10 - l into the field equation:

A = l(10 - l) = 10l - l²

Derivative and critical points: Differentiate A with respect to l.

A' = 10 - 2l

Set the derivative to zero to find the critical point:

10 - 2l = 0 l = 5

At l = 5, we find the critical point. Now, determine whether it is a maximum or a minimum by considering intervals or using the second derivative.

Second Derivative Test: Differentiate A' to find A''.

A'' = -2

Since A'' is less than zero, the function is concave downward, which indicates a maximum.

Therefore, the rectangle will have maximum area when l = 5, and by extension w = 10 - 5 = 5.

Hence a square with 5 unit sides gives a maximum area of 25 square units.

Visual example of a rectangle

A graphics representation example can be used to show how a rectangle behaves:

Length Width

Example: Minimizing costs

Suppose you want to make a cylindrical can that uses the least amount of material (cost) and can hold a certain amount of goods. Suppose the can can hold 500 cubic centimeters of goods.

Define the objective function: The cost probably depends on the surface area of the can, including the circular top and bottom as well as the sides.

Volume, V = πr²h = 500 Surface Area, S = 2πr² + 2πrh

Solve the volume equation for the height h:

h = 500 / (πr²)

Substitute the value of h into the surface area equation:

S = 2πr² + 2πr(500 / πr²) = 2πr² + 1000/r

Find the derivative: Differentiate S with respect to r.

S' = 4πr - 1000/r²

To find the critical points, set S' to zero:

4πr - 1000/r² = 0 4πr³ = 1000 r³ = 250/π r = (250/π)^(1/3)

Solve for h using the value of r. This will give the dimensions that will minimize the surface area.

Second derivative test

To use the second derivative test, find S'' :

S'' = 4π + 2000/r³

Since S'' is positive, it indicates the minimum surface area at the critical point.

Draw conclusions, obtaining optimal dimensions with the calculated radius and height.

Final thoughts on optimization problems

Understanding optimization is important for solving real-world problems. The essence of optimization is to find the most effective solution under given constraints. Whether you are determining the best use of materials or aiming for maximum efficiency, methods learned through calculus can be applied to achieve these goals.

Finally, practicing a variety of optimization problems strengthens understanding and proficiency in applying calculus concepts to practical scenarios.


Grade 11 → 4.3.5


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Optimization Problems

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