Allied & ICAR Exam Bank • ALLIED_0780 NAT • 2 Marks

Section 1: Engineering Mathematics

Calculus

Subtopic: Maxima and minima

Question Statement:

The function $f(x, y) = x^2 + y^2 - 4x - 6y + 13$ has a local minimum at $(x_0, y_0)$. The minimum value of $f(x, y)$ is:

Answer Key:

0

Step-by-Step Solution & Explanation

$f(x, y) = (x - 2)^2 - 4 + (y - 3)^2 - 9 + 13 = (x - 2)^2 + (y - 3)^2$. The minimum value is $0$ at $(2, 3)$.
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