Question

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

• A. $12$
• B. $24$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is B

$24$

Explanation for the correct option:

Find the minimum value of $ab$:

Given, $ab=2a+3b$

$\Rightarrow$$(a–3)b=2a$

$\Rightarrow$ $b=\frac{2a}{\left(a–3\right)}$

Let $z=ab$

$$\begin{gathered} =a\times \frac{2a}{\left(a-3\right)} \\ =\frac{2a^2}{\left(a–3\right)} \end{gathered}$$

Step 2. Differentiate it with respect to $a$:

$\begin{array}{rcl}\frac{dz}{da}& =& \frac{2\left[(a–3)2a–a^2\right]}{{(a–3)}^2}\\ & =& \frac{2\left(a^2–6a\right)}{{(a–3)}^2}\end{array}$

Put $\frac{dz}{da}=0$

$\Rightarrow$$\begin{array}{rcl}& & \frac{2\left(a^2–6a\right)}{{(a–3)}^2}\end{array}=0$

$\Rightarrow$ $a^2–6a=0$

$\Rightarrow$ $a=6$

Step 3. At $a=6$, $\frac{d^{2}z}{d{a}^2}=$ positive

when $a=6$ and $b=4$.

$\begin{array}{rcl}{\left(ab\right)}_{min}& =& 6\times 4\\ & =& \mathbf{24}\end{array}$

Hence, Option ‘B’ is Correct.