How many practice questions are available for Optimization?

SparkEd Math offers 60+ practice questions for Optimization across 3 difficulty levels (Easy, Medium, Hard) plus a timed test mode. Each question comes with visual step-by-step solutions and AI coaching hints.

Is Optimization content free on SparkEd Math?

Yes! All Optimization content on SparkEd Math is 100% free, including practice questions, visual solutions, lessons, and AI coaching. No credit card required.

How does SparkEd Math teach Optimization differently from textbooks?

SparkEd Math uses visual step-by-step solutions for every Optimization problem, making complex concepts easy to understand. The platform includes 3 difficulty levels for progressive learning, a Super Power Help AI coach for instant doubt-clearing, and interactive practice rather than passive reading.

Is SparkEd's Optimization content aligned with the IB syllabus?

Yes, all Optimization questions and lessons on SparkEd Math are fully aligned with the International Baccalaureate (MYP) (IB) syllabus for Class 10. The content follows the official curriculum structure and exam pattern.

Optimization — Class 10 IB (2025-26)

About Optimization — Class 10 IB

Solve optimization problems using algebraic and graphical methods in real-world contexts. This topic is part of the IB Class 10 mathematics syllabus (chapter: Unit 9). On this page you can practice 60 questions across three difficulty levels — 20 easy, 20 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 35-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Optimization — solved examples for Class 10 IB

Example 1easy

In the context of mathematical optimization, what does it mean to "optimize" a quantity?

A)A) To find a single possible value for that quantity.
B)B) To find the maximum or minimum possible value of that quantity under given conditions.
C)C) To simplify the algebraic expression representing that quantity.
D)D) To graph the quantity and analyze its shape.

Step-by-step solution

Optimization is a process of finding the 'best' solution from all feasible solutions.
This 'best' solution corresponds to either the maximum or minimum value of an objective function.
The specific goal (e.g., maximize profit, minimize cost) defines whether a maximum or minimum is sought.

Answer: B) To find the maximum or minimum possible value of that quantity under given conditions.

Example 2medium

A farmer wants to fence a rectangular plot of land adjacent to a river. No fencing is needed along the river. If the farmer has 200 meters of fencing, what is the maximum area of land that can be enclosed?

A)5000 m²
B)10000 m²
C)20000 m²
D)2500 m²

Step-by-step solution

Let the width of the rectangular plot (perpendicular to the river) be 'x' meters and the length (parallel to the river) be 'y' meters.
The total fencing available is 200 meters, so 2x + y = 200. This implies y = 200 - 2x.
The area of the plot is A = x × y. Substitute y: A(x) = x(200 - 2x) = 200x - 2x².
This is a quadratic function A(x) = -2x² + 200x. The maximum occurs at x = -b/(2a) = -200/(2 × -2) = 50. The maximum area is A(50) = 200(50) - 2(50)² = 10000 - 5000 = 5000 m².

Answer: 5000 m²

Example 3hard

A farmer has 200 meters of fencing and wants to enclose a rectangular field adjacent to a straight river. No fencing is needed along the river. What is the maximum area (in m²) of the field that the farmer can enclose?

A)5000 m²
B)4500 m²
C)4000 m²
D)5500 m²

Step-by-step solution

Let the width of the rectangular field perpendicular to the river be 'w' meters, and the length parallel to the river be 'l' meters. The total fencing used is 2w + l = 200 m.
From the fencing constraint, l = 200 - 2w. The area of the field, A, is given by A = l × w.
Substitute 'l' into the area formula: A(w) = (200 - 2w)w = 200w - 2w². This is a downward-opening parabola.
The maximum area occurs at the vertex of the parabola. The w-coordinate of the vertex is w = -b / (2a) = -200 / (2 × -2) = -200 / -4 = 50 m. The maximum area is A(50) = 200(50) - 2(50)² = 10000 - 2(2500) = 10000 - 5000 = 5000 m².

Answer: 5000 m²

Practice questions on Optimization

Q1.easy
A company wants to design a cylindrical can to hold 1 litre of liquid while using the least amount of material. What is the objective function they are trying to minimize?
1. A)A) The volume of the cylinder.
2. B)B) The height of the cylinder.
3. C)C) The surface area of the cylinder.
4. D)D) The radius of the cylinder.
Show answer
Answer: C) The surface area of the cylinder.
Hint: The amount of material used directly relates to the outer covering of the can.
Q2.easy
A student is tasked with finding two positive numbers whose sum is 20 and whose product is as large as possible. Which of the following correctly identifies a constraint in this problem?
1. A)A) The product of the numbers must be maximized.
2. B)B) One number must be greater than the other.
3. C)C) The numbers must be integers.
4. D)D) Both numbers must be greater than zero.
Show answer
Answer: D) Both numbers must be greater than zero.
Hint: Constraints are the limitations or conditions that the variables in the problem must satisfy.
Q3.easy
The graph of a quadratic function representing the height of a projectile over time is a parabola opening downwards. At what point on the graph does the maximum height of the projectile occur?
1. A)A) At the x-intercepts of the parabola.
2. B)B) At the vertex of the parabola.
3. C)C) At the y-intercept of the parabola.
4. D)D) At any point where the slope is zero.
Show answer
Answer: B) At the vertex of the parabola.
Hint: For a downward-opening parabola, the highest point is always its turning point.
Q4.medium
What is the minimum value of the function f(x) = 3x² - 12x + 15?
1. A)1
2. B)3
3. C)5
4. D)7
Show answer
Answer: 3
Hint: For a quadratic function ax² + bx + c where a > 0, the minimum value occurs at x = -b/(2a).
Q5.medium
Consider a quadratic function P(x) = ax² + bx + c that models the profit of a company. Which statement must be true for the company to achieve a maximum profit?
1. A)A) The value of 'a' must be positive.
2. B)B) The value of 'a' must be negative.
3. C)C) The value of 'b' must be zero.
4. D)D) The discriminant b² - 4ac must be negative.
Show answer
Answer: B) The value of 'a' must be negative.
Hint: The sign of the leading coefficient 'a' determines the direction in which the parabola opens.
Q6.medium
The height h (in meters) of a projectile launched upwards is given by the function h(t) = -5t² + 40t + 5, where t is the time in seconds. Which statement best describes the maximum height of the projectile based on its graph?
1. A)A) The maximum height is the y-intercept of the graph.
2. B)B) The maximum height is the point where the graph intersects the x-axis.
3. C)C) The maximum height is the vertex of the parabola, where the tangent line is horizontal.
4. D)D) The maximum height is found by setting h(t) = 0 and solving for t.
Show answer
Answer: C) The maximum height is the vertex of the parabola, where the tangent line is horizontal.
Hint: The path of a projectile under gravity is typically a parabolic curve.
Q7.hard
A square sheet of metal with sides 30 cm is used to make an open-top box by cutting out equal squares from each corner and folding up the sides. What is the maximum possible volume (in cm³) of the box?
1. A)1800 cm³
2. B)2000 cm³
3. C)2250 cm³
4. D)2500 cm³
Show answer
Answer: 2000 cm³
Hint: Let 'x' be the side length of the cut-out squares. Express the dimensions of the resulting box and then its volume in terms of 'x'. Consider the domain for 'x'.
Q8.hard
A quadratic function f(x) = ax² + bx + c models the profit of a company, where 'x' is the number of units produced. If 'a' is a negative real number, which of the following statements about the function and its graph is always true regarding the maximum profit?
1. A)The maximum profit is achieved when 'x' is the x-intercept of the parabola.
2. B)The graph of the function opens upwards, indicating a minimum profit.
3. C)The maximum profit corresponds to the y-coordinate of the vertex of the parabola.
4. D)The maximum profit can be found by setting f(x) = 0 and solving for 'x'.
Show answer
Answer: The maximum profit corresponds to the y-coordinate of the vertex of the parabola.
Hint: Recall the properties of a quadratic function when the leading coefficient 'a' is negative. What shape does the graph take, and where is its highest point?
Q9.hard
The height 'h' (in meters) of a projectile launched upwards is given by the function h(t) = -5t² + vt + c, where 't' is the time in seconds, 'v' is the initial velocity, and 'c' is the initial height. If the projectile reaches a maximum height of 45 meters after 3 seconds from an initial height of 0 meters, what was its initial velocity 'v' (in m/s)?
1. A)20 m/s
2. B)25 m/s
3. C)35 m/s
4. D)30 m/s
Show answer
Answer: 30 m/s
Hint: The maximum height occurs at the vertex of the parabolic path. Use the formula for the time at which the vertex occurs.

These are 9 of the 60 questions available for Optimization. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.