The minimum value of 2x + 3y, when xy = 6, is | vedclass

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(A) Let $f(x, y) = 2x + 3y$ subject to the constraint $xy = 6.$Since $xy = 6,$ we have $y = \frac{6}{x}.$Substituting this into the function,we get $f

Vedclass · Accepted Answer

$12$

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(A) Let $f(x, y) = 2x + 3y$ subject to the constraint $xy = 6.$Since $xy = 6,$ we have $y = \frac{6}{x}.$Substituting this into the function,we get $f(x) = 2x + 3\left(\frac{6}{x}\right) = 2x + \frac{18}{x}.$To find the minimum,we find the derivative $f'(x) = 2 - \frac{18}{x^2}.$Setting $f'(x) = 0,$ we get $2 = \frac{18}{x^2},$ which implies $x^2 = 9,$ so $x = \pm 3.$Since $x$ and $y$ must be positive for a minimum value in this context (as $xy=6$),we take $x = 3.$Then $y = \frac{6}{3} = 2.$The minimum value is $f(3) = 2(3) + 3(2) = 6 + 6 = 12.$

The minimum value of $2x + 3y,$ when $xy = 6,$ is

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