Find the greatest value of `x^2 y^3`, where `x` and` y` lie in the first quadrant on the line `3x+4y=5`.

Options

A. `(3)/(4)`
B. `(3)/(8)`
C. `(3)/(16)`
D. `(1)/(16)`

Answer

C

Explanation

Let `P=x^(2)y^(3)`. Clearly, P is the product of 5 factors such that two of them are equal to x and the remaining 3 are equal to y.
Now,
`3x4y=5`
`implies2((3x)/(2))+3((4y)/(3))=5implies(3x)/(2)+(3x)/(2)+(4y)/(3)+(4y)/(3)+(4y)/(3)=5`
It is evident from this that the sum of five numbers such that two of them are equal to `(3x)/(2)` and each of the remaining three is equal to `(4y)/(3),` is constant equal to 5. So, their product
`P'=((3x)/(2))^(2)((4y)/(3))^(3)=((3)/(2))^(2)((4)/(3))^(3)x^(2)y^(3)=((3)/(2))^(2)((4)/(3))^(3)P" "...(i)`
will be maximum, if
`(3x)/(2)=(4y)/(3)`
`implies" "(3x)/(2)=(4y)/(3)=(3x+4y)/(5)`
`implies" "(3x)/(2)=(4y)/(3)=1" "[because3x+4y=5]`
`implies" "x=(2)/(3)" and "y=(3)/(4)`
Also from (i), P is maximum when P' is maximum.
Thus, P is maximum when `x=(2)/(3)" and "y=(3)/(4)`
The maximum value of `P i.e. x^(2)y^(3)" is "((2)/(3))^(2)((3)/(4))^(3)=(3)/(16)`