kmaheshkumar

• May 11th, 2007

 

Given
X^3*y^2=6
Y sq is positive=> X cube is positive=>X is positive=>Y is positive

Factors of 6 are 1,2,3,6
Values for X cube =1,2,3,6
Values for Y square=6,3,2,1

The least values are X=root cube 3 and Y=sqroot 2, since 3x+4y
Thus ,
Minimum value of 3X+4Y is 3*cuberoot3 + 4*sqroot2

Hence the Ans