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  4. If ab=2a+3b,a>0,b>0 , then find the minimum value of ab .

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Q. If $ab=2a+3b,a>0,b>0$ , then find the minimum value of $ab$ .

NTA AbhyasNTA Abhyas 2022

Solution:

$ab=2a+3b\Rightarrow b=\frac{2 a}{a - 3}$
Now, $z=ab=\frac{2 a^{2}}{a - 3}$
So, $\frac{d z}{d a}=\frac{2 \left(a^{2} - 6 a\right)}{\left(a - 3\right)^{2}}=0\Rightarrow a=0,6$
At $a=6,\frac{d^{2} z}{d a^{2}}=+ve$ So, $a=6,b=4$
$\therefore \left(a b\right)_{\min}=6\times 4=24$