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Optimization Notes for GATE DA

GATE DA Optimization: 1 chapters, 1 previous year questions (8% of Calculus and Optimization), 42 practice questions and one solved question from each chapter

A question from this chapter

Question 1

Level 3: Exam Standard

Find the maximum value of the function f (x, y) = x^{3} - 6 x^{2} + 9 x + y^{2} over the closed rectangular domain D = {(x, y) : 0 \leq x \leq 4, - 2 \leq y \leq 2} .

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Optimization Notes for GATE DA

GATE DA Optimization: 1 chapters, 1 previous year questions (8% of Calculus and Optimization), 42 practice questions and one solved question from each chapter.

About Optimization Notes

Full study notes for Optimization in GATE DA, organised across 1 chapters. Each chapter page explains concepts from the basics with worked examples and the formulas you need.

Optimization Weightage in GATE DA

Optimization accounts for 1 of 13 Calculus and Optimization previous year questions in our bank (8%), about 1 per paper across 1 papers.

Optimization Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Optimization: Local and Global Extrema	Local and Global Extrema in Optimization	1	100%	42

Optimization: Local and Global Extrema Notes (42 questions)

One Solved Question from Each Optimization Chapter

Question 1 · Optimization: Local and Global Extrema MCQ

Find the maximum value of the function $f (x, y) = x^{3} - 6 x^{2} + 9 x + y^{2}$ over the closed rectangular domain $D = {(x, y) : 0 \leq x \leq 4, - 2 \leq y \leq 2}$ .

A.
0
B.
4
C.
6
D.
8

Correct Answer:

Step-by-Step Solution

Key idea: To find the global maximum on a closed bounded domain, you must check both the interior stationary points and the boundaries of the domain. The trap is assuming the interior local maximum is the global maximum.

Step 1: Find interior stationary points.

$f_{x} = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3) = 0 ⟹ x = 1, 3$ .

$f_{y} = 2 y = 0 ⟹ y = 0$ .

Points: $(1, 0)$ and $(3, 0)$ .

$f (1, 0) = 1 - 6 + 9 + 0 = 4$ . (Local maximum).

$f (3, 0) = 27 - 54 + 27 + 0 = 0$ . (Local minimum).

Step 2: Check the boundaries.

Boundary $y = 2$ : $f (x, 2) = x^{3} - 6 x^{2} + 9 x + 4$ .

Derivative is $3 x^{2} - 12 x + 9 = 0 ⟹ x = 1, 3$ .

$f (1, 2) = 4 + 4 = 8$ .

$f (3, 2) = 0 + 4 = 4$ .

Endpoints: $f (0, 2) = 4$ , $f (4, 2) = 64 - 96 + 36 + 4 = 8$ .

Boundary $y = - 2$ : Same as $y = 2$ since $y^{2} = 4$ . Max is 8.

Boundary $x = 0$ : $f (0, y) = y^{2}$ . Max at $y = \pm 2$ , value 4.

Boundary $x = 4$ : $f (4, y) = 64 - 96 + 36 + y^{2} = 4 + y^{2}$ . Max at $y = \pm 2$ , value 8.

Step 3: Compare all values. Interior max is 4. Boundary max is 8.

The global maximum is 8.

Answer: 8

A question from this chapter

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Unlock the whole course

Optimization Notes for GATE DA

About Optimization Notes

Optimization Weightage in GATE DA

Optimization Chapter Matrix

More from Calculus and Optimization

One Solved Question from Each Optimization Chapter

Step-by-Step Solution