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.