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

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


2 Answers

suryakanth AskiitiansExpert-IITB
105 Points
12 years ago

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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..


Askiitians Expert

Suryakanth –IITB


vikas askiitian expert
509 Points
12 years ago

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...

Think You Can Provide A Better Answer ?