Explanation

(b)
Given,
$ab = 2a + 3b\ \ \Rightarrow \ \ (a - 3)b = 2a\ \ \Rightarrow \ \ b = \frac{2a}{a - 3}$
Now, let $\mathcal{z =}ab = \frac{2a^{2}}{a - 3}$
On differentiating w.r.t. $x$, we get
$\frac{d\mathcal{z}}{sa} = \frac{2\lbrack(a - 3)3a - a^{2}\rbrack}{(a - 3)^{2}} = \frac{2\lbrack a^{2} - 6a\rbrack}{(a - 3)^{2}}$
For a minimum, put $\frac{d\mathcal{z}}{da} = 0$
$\Rightarrow \ \ \ a^{2} - 6a = 0\ \ \Rightarrow \ \ a = 0,\ 6$
At $a = 6,\ \ \frac{d^{2}\mathcal{z}}{da^{2}} = + ve$
When, $a = 6,\ \ b = 4$
$\therefore\ (ab)_{\min} = 6 \times 4 = 24$