Question

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

Answer 1

Answer: 24

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