Question

If $ab = 2a + 3b$ , $a > 0$ , $b > 0$ then the minimum value of $ab$ is
a. $12$
b. $24$
c. $\dfrac{1}{4}$
d. None of these

Answer 1

Hint : Here in this question $ab = 2a + 3b$ by re-altering and we are going to solve. Let we consider the $ab = z$ and we will apply differentiation so we can obtain the result. Here we will find the minimum value of the product of two numbers.

Complete step-by-step answer :
Consider the given data that is $ab = 2a + 3b$ . Move 3b to the LHS we have
$\Rightarrow ab - 3b = 2a$
Take b as common on LHS and we can rewrite the equation as
$\Rightarrow (a - 3)b = 2a$
Taking $(a - 3)$ on RHS and the b can be rewritten as
$\Rightarrow b = \dfrac{{2a}}{{(a - 3)}}$
Now we will consider $ab = z$ , by substituting the value of b we have
$\Rightarrow z = a \cdot \dfrac{{2a}}{{(a - 3)}}$
By multiplying,
$\Rightarrow z = \dfrac{{2{a^2}}}{{(a - 3)}}$
Differentiate the above equation w.r.t, a we have
$\dfrac{{dz}}{{da}} = \dfrac{d}{{da}}\left( {\dfrac{{2{a^2}}}{{(a - 3)}}} \right)$
To differentiate we apply the quotient rule that is $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ so we have,
$\Rightarrow \dfrac{{dz}}{{da}} = \dfrac{{(a - 3)\dfrac{d}{{da}}(2{a^2}) - 2{a^2}\dfrac{d}{{da}}(a - 3)}}{{{{(a - 3)}^2}}}$
On differentiation we have
$\Rightarrow \dfrac{{dz}}{{da}} = \dfrac{{(a - 3)4a - 2{a^2}}}{{{{(a - 3)}^2}}}$
On further simplification
$\Rightarrow \dfrac{{dz}}{{da}} = \dfrac{{4{a^2} - 12a - 2{a^2}}}{{(a - 3)}}$
$\Rightarrow \dfrac{{dz}}{{da}} = \dfrac{{2{a^2} - 12a}}{{{{(a - 3)}^2}}}$
Since ab value will be constant, we can take differentiation of z has zero. Since the differentiation of constant function is zero.
We take $\Rightarrow \dfrac{{dz}}{{da}} = 0$
Therefore, we have $0 = \dfrac{{2{a^2} - 12a}}{{{{(a - 3)}^2}}}$
$\Rightarrow 2{a^2} - 12a = 0$
Divide the above equation by 2 we have
$\Rightarrow {a^2} - 6a = 0$
$\Rightarrow a(a - 6) = 0$
Hence, we have $a = 0$ or $a = 6$
In the question they have mentioned that a is greater than 0 and b is greater than 0 that is $a > 0$ , $b > 0$ so we are considering the value of a has 6
By substituting the value of a in $b = \dfrac{{2a}}{{(a - 3)}}$ we have
$\Rightarrow b = \dfrac{{2(6)}}{{(6 - 3)}}$
On simplification
$\Rightarrow b = \dfrac{{12}}{3}$
Hence, we have $\Rightarrow b = 4$
Therefore, we have $a = 6$ and $b = 4$
The product of ab is $ab = (6)(4)$
Therefore $\Rightarrow ab = 24$
The minimum of $ab$ is 24
So, the correct answer is “Option b”.

Note : We can also answer the above question using arithmetic mean (A.M) and geometric mean (G.M). We have relation between the arithmetic mean and geometric mean that is $A.M > G.M$ , where $A.M = \dfrac{{a + b}}{2}$ and $G.M = \sqrt {ab}$ . By using this we can obtain the result.