If ab = 2a + 3b where a 0 and b 0,then th | vedclass

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(B) Given $ab = 2a + 3b$. Since $a > 0$ and $b > 0$,we can express $b$ in terms of $a$:$b(a - 3) = 2a \implies b = \frac{2a}{a - 3}$.Since $b > 0$ and

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$24$

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(B) Given $ab = 2a + 3b$. Since $a > 0$ and $b > 0$,we can express $b$ in terms of $a$:$b(a - 3) = 2a \implies b = \frac{2a}{a - 3}$.Since $b > 0$ and $a > 0$,we must have $a - 3 > 0$,so $a > 3$.Let $z = ab = a \left( \frac{2a}{a - 3} ight) = \frac{2a^2}{a - 3}$.To find the minimum,we differentiate $z$ with respect to $a$:$\frac{dz}{da} = \frac{(a - 3)(4a) - 2a^2(1)}{(a - 3)^2} = \frac{4a^2 - 12a - 2a^2}{(a - 3)^2} = \frac{2a^2 - 12a}{(a - 3)^2}$.Setting $\frac{dz}{da} = 0$ gives $2a(a - 6) = 0$. Since $a > 3$,we have $a = 6$.At $a = 6$,$b = \frac{2(6)}{6 - 3} = \frac{12}{3} = 4$.The value of $ab = 6 	imes 4 = 24$.Using the second derivative test or checking values around $a=6$,we confirm this is the minimum value.Thus,the minimum value of $ab$ is $24$.

If $ab = 2a + 3b$ where $a > 0$ and $b > 0$,then the minimum value of $ab$ is:

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