Question Bank • QB_EXP_EM_PD_020 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 $f(x, y) = x^2 + y^2 - 42x - 46y$ on $\mathbb{R}^2$ is equal to ________ (answer in integer).

Answer Key:

-970

Step-by-Step Solution & Explanation

Completing the square for $x$ and $y$: $f(x, y) = (x^2 - 42x + 441) + (y^2 - 46y + 529) - (441 + 529)$ $f(x, y) = (x - 21)^2 + (y - 23)^2 - (970)$ Since $(x - 21)^2 \ge 0$ and $(y - 23)^2 \ge 0$, the minimum occurs at $(x, y) = (21, 23)$: $f_{\min} = -(970) = -970$
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