Discussion
The maximum of \( f = xy \) subject to \( x + y = 10 \) is
Answer: D
With \( \nabla f = \lambda \nabla g \) we get \( y = \lambda \) and \( x = \lambda \), so \( x = y \). The constraint then gives \( x = y = 5 \) and
\[ f_{\max} \;=\; xy \;=\; 5 \times 5 \;=\; 25. \]
Substituting \( y = 10 - x \) and maximising \( 10x - x^{2} \) confirms the same vertex, so a fixed perimeter is enclosed most efficiently by the square
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