Question Bank • QB_EXP_EM_PD_017
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 $f(x, y) = x^2 + y^2 - 36x - 40y$ on $\mathbb{R}^2$ is equal to ________ (answer in integer).
Step-by-Step Solution & Explanation
Completing the square for $x$ and $y$:
$f(x, y) = (x^2 - 36x + 324) + (y^2 - 40y + 400) - (324 + 400)$
$f(x, y) = (x - 18)^2 + (y - 20)^2 - (724)$
Since $(x - 18)^2 \ge 0$ and $(y - 20)^2 \ge 0$, the minimum occurs at $(x, y) = (18, 20)$:
$f_{\min} = -(724) = -724$
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