Optimization: Local and Global Extrema Practice Questions for GATE DA

Solve 42+ Optimization: Local and Global Extrema practice questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Optimization and Extrema

Chapter Roadmap

The Optimization Journey

The Terrain

Local neighborhoods vs global domain. Strict vs non-strict extrema.

Finding Flat Spots

Using the gradient to locate stationary points where slope is zero.

Reading the Curvature

Deploying the Hessian matrix. Classifying via eigenvalues and principal minors.

The Convexity Shortcut

Proving that for convex functions, every local minimum is the global minimum.

The Core Goal of Optimization

Core Concept

The Objective

Optimization is the mathematical framework for finding the best value of an objective function $f (x)$ over a feasible set.

Minimization

Find

x^{*}

such that

f (x^{*}) \leq f (x)

for all valid

x

Loss functions in ML

Maximization

Find

x^{*}

such that

f (x^{*}) \geq f (x)

Likelihood functions

Key Insight

Maximizing

f (x)

is identical to minimizing

- f (x)

. We only need tools for minimization.

Optimization: Local and Global Extrema: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Calculus and Optimization MCQ

Consider the following assertion and reason: Assertion (A): The function

f (x, y) = x^{2} - y^{2}

has a local minimum at

(0, 0)

. Reason (R): The Hessian matrix at

(0, 0)

(20 0 - 2)

, which has eigenvalues

2

and

- 2

. Which of the following is correct?

A.
Both A and R are true, and R is the correct explanation of A
B.
Both A and R are true, but R is NOT the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Key idea: The second-order sufficient condition classifies stationary points based on the definiteness of the Hessian.

Step 1: Verify the stationary point.

$\nabla f = (2 x, - 2 y) = (0, 0) ⟹ (0, 0)$ is a stationary point.

Step 2: Compute the Hessian at $(0, 0)$ .

$H = (\frac{\partial ^{2} f}{\partial x ^{2}} \frac{\partial ^{2} f}{\partial y \partial x} \frac{\partial ^{2} f}{\partial x \partial y} \frac{\partial ^{2} f}{\partial y ^{2}}) = (20 0 - 2)$

The eigenvalues are $2$ and $- 2$ . So Reason (R) is TRUE.

Step 3: Classify using the second-order condition.

Since the eigenvalues have mixed signs ( $2 > 0$ and $- 2 < 0$ ), the Hessian is indefinite.

An indefinite Hessian at a stationary point indicates a saddle point, NOT a local minimum.

Therefore, Assertion (A) is FALSE.

Answer: A is false but R is true

Question 2 · Calculus and Optimization MCQ

Which of the following scenarios is IMPOSSIBLE for a strictly convex function $f : R^{2} \to R$ ?

A.
The function has exactly one local minimum
B.
The function has no stationary points
C.
The function has a local maximum at some point
D.
The function has a global minimum

Correct Answer:

Step-by-Step Solution

Key idea: Strictly convex functions have strong restrictions on their extrema.

Step 1: Recall properties of strictly convex functions.

The Hessian $H (x)$ is positive definite everywhere
Any local minimum is also a global minimum
There is at most one local (and global) minimum
The function cannot have any local maximum

Step 2: Analyze each option.

A) "Exactly one local minimum" — POSSIBLE. If a minimum exists, it's unique for strictly convex functions.

B) "No stationary points" — POSSIBLE. Example: $f (x, y) = e^{x} + e^{y}$ is strictly convex ( $H = diag (e^{x}, e^{y}) ≻ 0$ ) but $\nabla f = (e^{x}, e^{y}) \neq = 0$ everywhere.

C) "Local maximum at some point" — IMPOSSIBLE. At a local maximum, $H$ would need to be negative semi-definite, but for strictly convex functions, $H ≻ 0$ everywhere. Contradiction.

D) "Global minimum" — POSSIBLE. Example: $f (x, y) = x^{2} + y^{2}$ is strictly convex and has a global minimum at $(0, 0)$ .

Answer: The function has a local maximum at some point

Question 3 · Calculus and Optimization MCQ

Consider the following assertion and reason: Assertion (A): The function

f (x, y) = x^{2} + 2 y^{2} - 2 x y - 2 y

has a strict local minimum at

(1, 1)

. Reason (R): The Hessian matrix at

(1, 1)

(2 - 1 - 1 4)

, which has a positive determinant. Which of the following is correct?

A.
Both A and R are true, and R is the correct explanation of A
B.
Both A and R are true, but R is NOT the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Correct Answer:

Step-by-Step Solution

Key idea: The second-order sufficient condition requires the correct computation of the Hessian matrix, specifically the mixed partial derivatives.

Step 1: Verify the stationary point and Assertion (A).

$f_{x} = 2 x - 2 y = 0 ⟹ x = y$

$f_{y} = 4 y - 2 x - 2 = 0 ⟹ 4 x - 2 x - 2 = 0 ⟹ 2 x = 2 ⟹ x = 1, y = 1$ .

So $(1, 1)$ is indeed a stationary point.

Step 2: Compute the true Hessian to check Reason (R).

$f_{xx} = 2$

$f_{y y} = 4$

$f_{x y} = \frac{\partial}{\partial y} (2 x - 2 y) = - 2$

The true Hessian is $(2 - 2 - 2 4)$ .

Step 3: Evaluate the statements.

True $det (H) = (2) (4) - (- 2)^{2} = 8 - 4 = 4 > 0$ .

Since $f_{xx} = 2 > 0$ and $det > 0$ , it is a strict local minimum. Assertion (A) is TRUE.

However, Reason (R) claims the off-diagonal entries are $- 1$ . This is a mismatch of the coefficient; the derivative of $- 2 x y$ with respect to both $x$ and $y$ is $- 2$ , not $- 1$ . Thus, Reason (R) is FALSE.

Answer: A is true but R is false

Question 4 · Calculus and Optimization MCQ

Let $f (x, y)$ be a twice continuously differentiable function such that its Hessian matrix $H (x, y)$ has non-negative eigenvalues ( $λ_{1} \geq 0, λ_{2} \geq 0$ ) at every point in $R^{2}$ . Which of the following features is IMPOSSIBLE for the graph of $f$ ?

A.
A flat valley where every point is a global minimum
B.
A unique stationary point that is a strict global minimum
C.
A saddle point where the function curves upwards in one direction and downwards in another
D.
No stationary points at all, with the function decreasing towards infinity

Correct Answer:

Step-by-Step Solution

Key idea: Non-negative eigenvalues everywhere mean the Hessian is positive semi-definite everywhere, which is the definition of a convex function.

Step 1: Recall the properties of convex functions.

If $H ⪰ 0$ everywhere, $f$ is convex. For convex functions:

Any local minimum is a global minimum.
The set of global minima is convex (can be a single point or a flat region/valley).
There are no saddle points, because a saddle point requires the Hessian to be indefinite (mixed positive and negative eigenvalues) at that point.

Step 2: Evaluate the options.

A) A flat valley: Possible. Example: $f (x, y) = (x - y)^{2}$ . $H = (2 - 2 - 2 2)$ , eigenvalues $0, 4 \geq 0$ . The line $y = x$ is a flat valley of global minima.

B) Unique strict global min: Possible. Example: $f (x, y) = x^{2} + y^{2}$ . $H = (2002)$ , eigenvalues $2, 2 > 0$ .

C) A saddle point: IMPOSSIBLE. A saddle point requires the Hessian to have at least one strictly negative eigenvalue to curve downwards in some direction. But we are given $λ_{1}, λ_{2} \geq 0$ everywhere.

D) No stationary points: Possible. Example: $f (x, y) = e^{- x} + e^{- y}$ . $H = (e^{- x} 0 0 e^{- y}) ≻ 0$ . Gradient is never zero.

Answer: A saddle point where the function curves upwards in one direction and downwards in another

Question 5 · Calculus and Optimization MCQ

For the function $f (x, y) = x^{2} + y^{2} - 2 x - 4 y + 5$ , a student finds stationary points by setting each partial derivative to zero independently and lists four candidates: $(1, 0)$ , $(0, 2)$ , $(1, 2)$ , and $(0, 0)$ . Which of these points is the actual location of the minimum?

A.
(0, 0)
B.
(1, 0)
C.
(0, 2)
D.
(1, 2)

Correct Answer:

Step-by-Step Solution

Key idea: The first-order necessary condition requires ALL partial derivatives to be zero simultaneously, not just one at a time.

Step 1: Compute the gradient.

$\frac{\partial f}{\partial x} = 2 x - 2$

$\frac{\partial f}{\partial y} = 2 y - 4$

Step 2: Set both components to zero simultaneously.

$2 x - 2 = 0 ⟹ x = 1$

$2 y - 4 = 0 ⟹ y = 2$

The only stationary point is $(1, 2)$ .

Step 3: Verify this is a minimum.

The Hessian is $H = (2002)$ , which is positive definite. Thus $(1, 2)$ is a strict local minimum.

The student's error was treating the system as two independent equations and listing all combinations. Only $(1, 2)$ satisfies BOTH equations simultaneously.

Answer: (1, 2)

Question 6 · Calculus and Optimization MCQ

A student finds the stationary points of $f (x, y) = x^{2} y - x^{2} + y^{2} - 4 y + 4$ by setting the factors of the partial derivatives to zero, yielding $x = 0$ or $y = 1$ from $f_{x} = 0$ , and $x^{2} + 2 y - 4 = 0$ from $f_{y} = 0$ . They incorrectly pair $x = 0$ and $y = 1$ to claim $(0, 1)$ is a stationary point. What is the minimum value of $f (x, y)$ among its ACTUAL stationary points?

A.
-2
B.
0
C.
1
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: The first-order necessary condition requires ALL partial derivatives to be zero simultaneously. Pairing roots from independent equations without verifying the full system leads to invalid candidates.

Step 1: Find the actual stationary points by solving the system:

$f_{x} = 2 x (y - 1) = 0 ⟹ x = 0 or y = 1$

$f_{y} = x^{2} + 2 y - 4 = 0$

Case 1: If $x = 0$ , substitute into $f_{y} = 0$ :

$0 + 2 y - 4 = 0 ⟹ y = 2$ . Point: $(0, 2)$ .

Case 2: If $y = 1$ , substitute into $f_{y} = 0$ :

$x^{2} + 2 (1) - 4 = 0 ⟹ x^{2} = 2 ⟹ x = \pm 2$ . Points: $(2, 1)$ and $(- 2, 1)$ .

Step 2: Classify the points using the Hessian.

$f_{xx} = 2 y - 2$ , $f_{y y} = 2$ , $f_{x y} = 2 x$ .

At $(0, 2)$ : $f_{xx} = 2$ , $f_{y y} = 2$ , $f_{x y} = 0$ .

$D_{1} = 2 > 0$ , $D_{2} = 4 - 0 = 4 > 0$ . Strict local minimum.

Value: $f (0, 2) = 0 - 0 + 4 - 8 + 4 = 0$ .

At $(\pm 2, 1)$ : $f_{xx} = 0$ , $f_{y y} = 2$ , $f_{x y} = \pm 22$ .

$D_{2} = 0 - 8 = - 8 < 0$ . Saddle points.

The only local minimum is at $(0, 2)$ with a value of $0$ .

Answer: 0

Question 7 · Calculus and Optimization MCQ

For $f (x, y) = x^{3} + y^{2} - 3 x$ , the stationary points are $(1, 0)$ and $(- 1, 0)$ . A student additionally claims $(0, 0)$ is a local minimum since $f (0, y) = y^{2} \geq 0$ for all $y$ . Given that $\nabla f (0, 0) = (- 3, 0) \neq = 0$ , this contradicts Fermat's theorem. Among the actual stationary points, the minimum value of $f$ is:

A.
-3
B.
-2
C.
0
D.
2

Correct Answer:

Step-by-Step Solution

Key idea: This is a Fermat's theorem question, recognisable because a non-stationary point is claimed as a local extremum. The necessary condition $\nabla f = 0$ rules it out immediately.

Step 1: The question states the stationary points are $(1, 0)$ and $(- 1, 0)$ . The student's extra candidate $(0, 0)$ is invalid because $\nabla f (0, 0) = (- 3, 0) \neq = 0$ . By Fermat's theorem, a differentiable function cannot have a local extremum at a point where the gradient is nonzero. The student's error was checking only along the $y$ -axis; along the $x$ -axis, $f (x, 0) = x^{3} - 3 x$ , which decreases for small positive $x$ .

Step 2: Evaluate $f$ at the two valid stationary points.

$f (1, 0) = 1^{3} + 0^{2} - 3 (1) = 1 - 3 = - 2$

$f (- 1, 0) = (- 1)^{3} + 0^{2} - 3 (- 1) = - 1 + 3 = 2$

Step 3: The smaller value is $- 2$ .

Answer: $- 2$

Question 8 · Calculus and Optimization MCQ

For a twice continuously differentiable function $f (x, y)$ , the Hessian matrix $H$ at any point must satisfy which of the following properties?

A.
$H$ is always positive semi-definite
B.
$H$ is a symmetric matrix
C.
All eigenvalues of $H$ are strictly positive
D.
$det (H) > 0$

Correct Answer:

Step-by-Step Solution

Key idea: The Hessian matrix is always symmetric for smooth functions due to equality of mixed partial derivatives.

Step 1: Recall the definition of the Hessian.

For $f (x, y)$ , the Hessian is:

$H = (\frac{\partial ^{2} f}{\partial x ^{2}} \frac{\partial ^{2} f}{\partial y \partial x} \frac{\partial ^{2} f}{\partial x \partial y} \frac{\partial ^{2} f}{\partial y ^{2}})$

Step 2: Apply Clairaut's theorem.

For twice continuously differentiable functions, mixed partial derivatives are equal:

$\frac{\partial ^{2} f}{\partial x \partial y} = \frac{\partial ^{2} f}{\partial y \partial x}$

Therefore, $H = H^{T}$ , meaning $H$ is symmetric.

Step 3: Check other options.

A is false: $H$ can be indefinite (e.g., saddle points)
C is false: eigenvalues can be negative or zero
D is false: $det (H)$ can be zero or negative

Answer: $H$ is a symmetric matrix

Question 9 · Calculus and Optimization MCQ

Match the functions in List I with their classification at the stationary point

(0, 0)

in List II: List I (Functions): P)

f (x, y) = x^{2} + y^{2}

f (x, y) = - x^{2} - y^{2}

f (x, y) = x^{2} - y^{2}

f (x, y) = x^{2} + y^{2} + 2 x y

List II (Classification at $(0, 0)$ ): 1) Saddle point 2) Local minimum 3) Local maximum 4) Inconclusive (semi-definite)

A.
P-2, Q-3, R-1, S-4
B.
P-2, Q-1, R-3, S-4
C.
P-3, Q-2, R-1, S-4
D.
P-2, Q-3, R-4, S-1

Correct Answer:

Step-by-Step Solution

Key idea: Compute the Hessian at $(0, 0)$ for each function and check definiteness.

Step 1: Analyze P) $f (x, y) = x^{2} + y^{2}$

$H = (2002)$ , eigenvalues $2, 2 > 0$ → positive definite → local minimum (2)

Step 2: Analyze Q) $f (x, y) = - x^{2} - y^{2}$

$H = (- 2 0 0 - 2)$ , eigenvalues $- 2, - 2 < 0$ → negative definite → local maximum (3)

Step 3: Analyze R) $f (x, y) = x^{2} - y^{2}$

$H = (20 0 - 2)$ , eigenvalues $2, - 2$ (mixed) → indefinite → saddle point (1)

Step 4: Analyze S) $f (x, y) = x^{2} + y^{2} + 2 x y = (x + y)^{2}$

$H = (2222)$ , $det (H) = 4 - 4 = 0$ → semi-definite → inconclusive (4)

Matching: P-2, Q-3, R-1, S-4

Answer: P-2, Q-3, R-1, S-4

Question 10 · Calculus and Optimization MCQ

Consider the following statements regarding the Hessian matrix $H$ of a twice differentiable function $f (x, y)$ at a stationary point, where $D = det (H)$ and $T = tr (H)$ :

P) If $D > 0$ and $T < 0$ , the point is a strict local maximum.

Q) If $D < 0$ , the point is a saddle point.

R) If $T < 0$ , the point is a strict local maximum.

Which of the above statements are ALWAYS true?

A.
P and Q only
B.
P and R only
C.
Q and R only
D.
P, Q, and R

Correct Answer:

Step-by-Step Solution

Key idea: The second derivative test in 2D relies on both the determinant (product of eigenvalues) and the trace (sum of eigenvalues) to determine definiteness.

Step 1: Analyze Statement P.

$D = λ_{1} λ_{2} > 0$ means eigenvalues have the same sign.

$T = λ_{1} + λ_{2} < 0$ means their sum is negative.

Therefore, both eigenvalues must be strictly negative ( $H$ is negative definite). This guarantees a strict local maximum. (P is TRUE).

Step 2: Analyze Statement Q.

$D = λ_{1} λ_{2} < 0$ means the eigenvalues have opposite signs.

This makes $H$ indefinite, which guarantees a saddle point. (Q is TRUE).

Step 3: Analyze Statement R.

$T = λ_{1} + λ_{2} < 0$ only tells us the sum is negative. It does not prevent one eigenvalue from being positive and the other being a larger negative number (e.g., $λ_{1} = 5, λ_{2} = - 10 ⟹ T = - 5, D = - 50$ ).

In this case, $D < 0$ , so it would be a saddle point, not a maximum. R ignores the constraint that $D$ must be positive. (R is FALSE).

Answer: P and Q only

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Optimization: Local and Global Extrema Practice Questions for GATE DA

Chapter Roadmap: Optimization and Extrema

The Optimization Journey

The Core Goal of Optimization

The Objective

Optimization: Local and Global Extrema: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution