medium difficulty12 min

Real-Life Optimisation Problems

Optimisation uses derivatives to find the maximum or minimum value of a quantity under given conditions or constraints.

In one line: Optimisation answer chain: choose the quantity, use the constraint, reduce to one variable, differentiate, test, then translate back to the story.

Practice This Concept

Understand it

A real-life optimisation problem is first a modelling problem and only then a differentiation problem.
The question usually gives a constraint, such as fixed perimeter, fixed volume, fixed surface area, or a relation between dimensions.
The quantity to be maximised or minimised must be written as an objective function in one variable, using the constraint to eliminate the other variable.
The feasible domain matters because lengths, radii, heights, areas, and costs cannot take arbitrary values.
After solving Q'(x) = 0, students must verify whether the critical value gives a maximum or minimum by using Q''(x), sign change, or endpoint comparison, and then answer the practical question with correct units.
Marks are often lost not in differentiation, but in choosing the wrong objective or forgetting to interpret the result.

Example

For a rectangle with perimeter 20 cm, let length be x and breadth be 10 - x. Area A = x(10 - x) = 10x - x². A' = 10 - 2x, so x = 5. Since A'' = -2 < 0, area is maximum when x = 5 and breadth = 5; the rectangle is a square.

Formula and rule

Optimisation method: form objective function Q; use constraint to express Q in one variable; solve Q'(x) = 0 within the feasible domain; verify maximum or minimum by second derivative test, sign test, or endpoint comparison where needed.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A rectangular sheet has perimeter 40 cm.

1Find the rectangle of maximum area.
2Let length be x cm and breadth be y cm.
3Constraint: 2x + 2y = 40, so y = 20 - x.
4Area A = xy = x(20 - x) = 20x - x², where 0 < x < 20.
5Differentiate: A'(x) = 20 - 2x.
6Set A'(x) = 0: 20 - 2x = 0, so x = 10.
7A''(x) = -2 < 0, so the area is maximum.
8Then y = 20 - 10 = 10.
9The rectangle of maximum area is 10 cm by 10 cm.

Method you can reuse

Read what is to be maximised or minimised; assign variables with units; write the constraint; express the objective in one variable; find derivative and critical value; test the result; interpret the answer in words with units.

Visual tip

A diagram may help for geometry-based word problems, but the essential requirement is a labelled variable model and constraint equation.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

A rectangle has perimeter 24 cm. What dimensions give maximum area?

Think first, then show answer

Let length x and breadth 12 - x. Area A = x(12 - x). A' = 12 - 2x = 0 gives x = 6, so breadth = 6. Maximum area occurs for a 6 cm by 6 cm square.

Common wrong answer

Finding x = 10 correctly in a rectangle problem but stopping there, even though the question asks for the maximum area or for both dimensions. Another realistic wrong answer is accepting a negative length or radius because the feasible domain was not checked.

Don't mix this up

A common error is forming the correct constraint but optimising the wrong expression, such as differentiating perimeter when the question asks for maximum area, or keeping two variables in the objective and treating one of them as constant. This leads to a neat-looking derivative but an answer that does not match the real situation.

Exam tip

Define variables clearly and mention the feasible domain, such as positive length, positive radius, or dimensions within the given constraint.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Usually appears as a long-answer or case-study problem involving maximum area, maximum volume, minimum cost, minimum surface area, or shortest distance.

1
Enough for 1 mark
Optimisation uses derivatives to find the maximum or minimum value of a quantity under given conditions or constraints.
2
Add for 2 marks
Example: For a rectangle with perimeter 20 cm, let length be x and breadth be 10 - x. Area A = x(10 - x) = 10x - x². A' = 10 - 2x, so x = 5. Since A'' = -2 < 0, area is maximum when x = 5 and breadth = 5; the rectangle is a square.
3
Add for 3 marks
A real-life optimisation problem is first a modelling problem and only then a differentiation problem.

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