Question Bank • QB_EXP_EM_PD_029
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 $f(x, y) = x^2 + y^2 - 60x - 64y$ on $\mathbb{R}^2$ is equal to ________ (answer in integer).
Step-by-Step Solution & Explanation
Completing the square for $x$ and $y$:
$f(x, y) = (x^2 - 60x + 900) + (y^2 - 64y + 1024) - (900 + 1024)$
$f(x, y) = (x - 30)^2 + (y - 32)^2 - (1924)$
Since $(x - 30)^2 \ge 0$ and $(y - 32)^2 \ge 0$, the minimum occurs at $(x, y) = (30, 32)$:
$f_{\min} = -(1924) = -1924$
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