Q: The greatest value of x^{2}y^{3} is, where x > 0 and y > 0 are connected by the relation 3x + 4y = 5

(A) 3/16

(B) 1/16

(C) 3/14

(D) none of these

Sol: (A). We are given 3x + 4y = 5 …(1)

Now, product x^{2}y^{3} will be maximum when $ (\frac{3}{2}x)^2 (\frac{4}{3}y)^3 $ is maximum. Above is product of 5 factors whose sum is

$\large 2.\frac{3}{2}x + 3.\frac{4}{3}y = 3x + 4y = 5 = constant $

Hence the product will be maximum when all the factors are equal

i.e. $\frac{3x}{2} = \frac{4 y}{3} = \frac{3x+4y}{2+3} = \frac{5}{5} = 1$

⇒ x = 2/3 , y = 3/4

∴ Maximum value of x^{2}y^{3} $\large = (\frac{2}{3})^2 (\frac{3}{4})^3 = \frac{3}{16} $