Find the minimum value of 2x + 3y when xy = 6 | vedclass

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(B) Given: $xy = 6$ $\implies y = \frac{6}{x}$ Let $f(x) = 2x + 3y = 2x + 3 \left( \frac{6}{x} \right) = 2x + \frac{18}{x}$ To find the minimum value,

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Given: $xy = 6$ $\implies y = \frac{6}{x}$ Let $f(x) = 2x + 3y = 2x + 3 \left( \frac{6}{x} \right) = 2x + \frac{18}{x}$ To find the minimum value,differentiate $f(x)$ with respect to $x$: $f'(x) = 2 - \frac{18}{x^2}$ Set $f'(x) = 0$ for critical points: $2 - \frac{18}{x^2} = 0 \implies x^2 = 9 \implies x = 3$ (assuming $x > 0$ for positive values) Now,find the second derivative: $f''(x) = \frac{36}{x^3}$ At $x = 3$,$f''(3) = \frac{36}{27} > 0$,which confirms a local minimum. The minimum value is $f(3) = 2(3) + \frac{18}{3} = 6 + 6 = 12$.

Find the minimum value of $2x + 3y$ when $xy = 6$.

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