Let $f ( x , y ) = x y$ with constraint $3 x ^ { 2 } + y ^ { 2 } = 6$ . Then f has a relative minimum at

A) $( 1 , \sqrt { 3 } )$
B) $( - 1 , \sqrt { 3 } )$
C) $( 1 , - \sqrt { 3 } )$
D) $( \pm 1 , \pm \sqrt { 3 } ) ( \pm 1 , \mp \sqrt { 3 } )$
E) $( \pm 1 , \mp \sqrt { 3 } )$

Question

Let $f ( x , y ) = x y$ with constraint $3 x ^ { 2 } + y ^ { 2 } = 6$ . Then f has a relative minimum at

A) $( 1 , \sqrt { 3 } )$
B) $( - 1 , \sqrt { 3 } )$
C) $( 1 , - \sqrt { 3 } )$
D) $( \pm 1 , \pm \sqrt { 3 } ) ( \pm 1 , \mp \sqrt { 3 } )$
E) $( \pm 1 , \mp \sqrt { 3 } )$

Quizplus · Accepted Answer

To find the relative minimum of $f(x, y) = xy$ subject to the constraint $3x^2 + y^2 = 6$, we can use the method of Lagrange multipliers. The gradient of $f$, $\nabla f = (y, x)$, must be parallel to the gradient of the constraint, $\nabla g = (6x, 2y)$, where $g(x, y) = 3x^2 + y^2 - 6$. Setting $\nabla f = \lambda \nabla g$, we get two equations: $y = 6\lambda x$ and $x = 2\lambda y$. Solving these equations along with the constraint gives the points where $f$ could have extrema. The correct points that satisfy both the constraint and the conditions for a minimum are $(\pm 1, \mp \sqrt{3})$, as these points are solutions to the system and represent where $f$ has its relative minima given the nature of the function and the constraint.

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