Question Bank • QB_EXP_EM_PD_011
NAT • 2 Marks
Section 1: Engineering Mathematics
Calculus: Partial Derivatives & Maxima-Minima
Subtopic: Constrained extrema and Lagrange multipliers
Question Statement:
The minimum value of the function $f(x, y) = x^2 + y^2 - 24x - 28y$ on $\mathbb{R}^2$ is equal to ________ (answer in integer).
Step-by-Step Solution & Explanation
Completing the square for $x$ and $y$:
$f(x, y) = (x^2 - 24x + 144) + (y^2 - 28y + 196) - (144 + 196)$
$f(x, y) = (x - 12)^2 + (y - 14)^2 - (340)$
Since $(x - 12)^2 \ge 0$ and $(y - 14)^2 \ge 0$, the minimum occurs at $(x, y) = (12, 14)$:
$f_{\min} = -(340) = -340$
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