Maxima and Minima of Multi-Variable Functions (Unconstrained)

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2023)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

This atom appeared once in 2023 as a 15-mark Section B computation. UPSC asks you to locate all critical points of a function of two variables and classify each as a local maximum, local minimum, or saddle point using the second-derivative test. The procedure is completely systematic, making this a reliable marks-source if you prepare it. Note the critical distinction from P1-CA-14 (Lagrange multipliers): here there is no constraint — you are optimising over all of $\mathbb{R}^2$.

Minimum Theory

Critical Points

A function $f(x, y)$ with continuous partial derivatives has a critical point at $(a, b)$ if

$f_x(a,b) = 0 \quad \text{and} \quad f_y(a,b) = 0.$

Every local extremum of a differentiable function is a critical point (necessary condition). Not every critical point is an extremum.

Second-Derivative Test (Hessian Criterion)

At a critical point $(a, b)$, compute the discriminant

$D = f_{xx}(a,b)\, f_{yy}(a,b) - \bigl[f_{xy}(a,b)\bigr]^2.$

Why $D$?

$D = \det H$ where $H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{pmatrix}$ is the Hessian matrix. $D > 0$ means $H$ is definite (sign determined by $f_{xx}$); $D < 0$ means $H$ is indefinite (saddle).

Functions of Three or More Variables

For $f(x, y, z)$: set all three first partials to zero. The Hessian is $3 \times 3$; classify using Sylvester’s criterion (all leading principal minors positive $\Rightarrow$ local min; alternating signs starting negative $\Rightarrow$ local max; otherwise saddle or inconclusive). UPSC questions typically stay with two variables.

Question Archetypes

find-and-classify-critical-points (1 question; 2023)

Recognition Cues

• A specific $f(x,y)$ is given — typically a polynomial or product of trigonometric/exponential terms
• The question asks for critical points and their nature (max/min/saddle)
• No constraint curve or region boundary is mentioned (distinguishes from Lagrange)
• 15 marks — full working (solving the critical-point system, computing $D$, classifying) expected

Solution Template

1. Compute $f_x$ and $f_y$.
2. Set $f_x = 0$ and $f_y = 0$; solve the system (often nonlinear — factor or substitute).
3. List all critical points.
4. For each: compute $f_{xx}$, $f_{yy}$, $f_{xy}$ at the point; evaluate $D = f_{xx}f_{yy} - f_{xy}^2$.
5. Apply the classification table; state the conclusion.
6. If $D = 0$, state that the test is inconclusive and attempt a direct comparison $f(a+h, b+k) - f(a,b)$ for small $(h,k)$.

Worked Example

2023 Paper 1, 2023-P1-Q4b (15 marks)

Find all the critical points of $f(x, y) = x^3 + y^3 - 3axy$ (where $a > 0$) and determine their nature.

Step 1. Compute first partial derivatives.

$f_x = 3x^2 - 3ay, \qquad f_y = 3y^2 - 3ax.$

Step 2. Set equal to zero.

$x^2 = ay \quad \cdots (1)$ $y^2 = ax \quad \cdots (2)$

From (1): $y = x^2/a$. Substitute into (2):

$\left(\frac{x^2}{a}\right)^2 = ax \implies \frac{x^4}{a^2} = ax \implies x^4 = a^3 x \implies x(x^3 - a^3) = 0.$

So $x = 0$ or $x = a$.

• $x = 0$: from (1), $y = 0$. Critical point: $(0, 0)$.
• $x = a$: from (1), $y = a^2/a = a$. Critical point: $(a, a)$.

Step 3. Compute second partial derivatives.

$f_{xx} = 6x, \qquad f_{yy} = 6y, \qquad f_{xy} = -3a.$

Step 4. Classify $(0, 0)$.

$f_{xx}(0,0) = 0, \quad f_{yy}(0,0) = 0, \quad f_{xy}(0,0) = -3a.$ $D = (0)(0) - (-3a)^2 = -9a^2 < 0 \quad (a > 0).$

$(0, 0)$ is a saddle point.

Step 5. Classify $(a, a)$.

$f_{xx}(a,a) = 6a, \quad f_{yy}(a,a) = 6a, \quad f_{xy}(a,a) = -3a.$ $D = (6a)(6a) - (-3a)^2 = 36a^2 - 9a^2 = 27a^2 > 0.$

Since $D > 0$ and $f_{xx} = 6a > 0$, the point $(a, a)$ is a local minimum.

The local minimum value is $f(a,a) = a^3 + a^3 - 3a \cdot a \cdot a = 2a^3 - 3a^3 = -a^3$.

$\boxed{(0,0) \text{ is a saddle point};\quad (a,a) \text{ is a local minimum with } f(a,a) = -a^3.}$

Common Traps

• Losing solutions when dividing: In Step 2, $x^4 = a^3 x$ — dividing both sides by $x$ loses the solution $x = 0$. Always factor instead.
• Forgetting to substitute back: After finding $x$, compute $y$ from the same equation — do not assume symmetry unless the system is symmetric.
• Sign of $f_{xy}$: $f_{xy}$ enters $D$ squared, so its sign does not matter for $D$, but an arithmetic error in $f_{xy}$ will flip the sign of $D$.
• Confusing with Lagrange: If a constraint like $g(x,y) = k$ appears, this procedure does not apply — that is P1-CA-14.

Marks-Aware Writing

For a 15-mark computation:

• 3 marks — correct $f_x$, $f_y$ computed and set to zero.
• 4 marks — critical-point system solved correctly, all critical points found (no solution dropped).
• 4 marks — second partial derivatives computed correctly at each critical point; $D$ evaluated.
• 4 marks — classification stated for each point with justification (citing the value of $D$ and $f_{xx}$); local extremum value stated if it is a max/min.

Always state $D$ explicitly and cite its sign — this is the key step that earns the classification marks.

Practice Set

Only one historical question on this atom (shown above).