GATE CS 1993 Q01.7 — Maxima and Minima, Mean Value Theorem, Integration

To find the local extrema of the function $f(x,y) = x^2y - 3xy + 2y + x$, we first find its critical points by setting the first-order partial derivatives to zero.

Step 1: Find the first-order partial derivatives

Step 2: Find the critical points

Set $f_y = 0$:

Set $f_x = 0$:

• For $x = 1$: $y = \frac{1}{3 - 2(1)} = 1$. Thus, $(1, 1)$ is a critical point.
• For $x = 2$: $y = \frac{1}{3 - 2(2)} = -1$. Thus, $(2, -1)$ is a critical point.

Step 3: Classify the critical points using the Second Derivative Test

Compute the second-order partial derivatives:

The discriminant $D(x,y)$ is given by:

Now, evaluate $D(x,y)$ at the critical points:

1. At $(1, 1)$: $$D(1, 1) = -(2(1) - 3)^2 = -1 < 0$$ Since $D < 0$, the point $(1, 1)$ is a saddle point.
2. At $(2, -1)$: $$D(2, -1) = -(2(2) - 3)^2 = -1 < 0$$ Since $D < 0$, the point $(2, -1)$ is also a saddle point.

Since both critical points are saddle points, the function $f(x,y)$ has no local extremum.