Question Bank • QB_EM_CALC_011
MCQ • 2 Marks
Section 1: Engineering Mathematics
Calculus
Subtopic: Maxima and minima of function with several independent variables
Question Statement:
For the function $f(x, y) = x^2 + y^2 + 6x - 4y + 13$, the critical point is:
Options:
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(A) Point of local maximum at $(-3, 2)$
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(B) Point of local minimum at $(-3, 2)$
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(C) Saddle point at $(3, -2)$
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(D) Point of local minimum at $(3, -2)$
Step-by-Step Solution & Explanation
Calculate first partial derivatives:
$\frac{\partial f}{\partial x} = 2x + 6 = 0 \implies x = -3$
$\frac{\partial f}{\partial y} = 2y - 4 = 0 \implies y = 2$
The critical point is $(-3, 2)$.
Second partial derivatives:
$r = \frac{\partial^2 f}{\partial x^2} = 2 > 0$
$s = \frac{\partial^2 f}{\partial x \partial y} = 0$
$t = \frac{\partial^2 f}{\partial y^2} = 2$
Discriminant $\Delta = rt - s^2 = (2)(2) - 0 = 4 > 0$.
Since $\Delta > 0$ and $r > 0$, $(-3, 2)$ is a point of local minimum.
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