GIven that 2x+3y=18, then find the maximum value of x^2y^3.

Dear arindam,

consider (2x/2)² (3y/3)³

Now sum of the factors of the above expression is

2(2x/2) + 3(3y/3) = 2x+3y

sum of the factors is constant (given 2x+3y = 18) then x² y³ is maximum when the factors are equal

=> 2x/2 = 3y/3 

^{=> x= y}

^{substuting in the given equation}

^{x = y = 18/5}

^{maximum value is = (18/5)^5}

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Suryakanth –IITB

let Z is a function whose maximum value we have to find..

     Z= x² y³

 2x +3y =18 (given).............1

   Z={ (18-3y)/2 }² y³    ...........2            (on sustituting x=18-3y/2 from eq 1)

  for maxima minima  put dZ/dy  =0  &

  find out different values of y ...put these values in eq 2 and at find the maximum value of this function...