It is given two positive real numbers `x,y` which are satisfied that `4x+3y=11`. Find the maximum value of the following function: `f(x,y)=(x+6)(y+7)(3x+2y)`

 It is given two positive real numbers `x,y` which are satisfied that `4x+3y=11`. 

Find the maximum value of the following function: 

`f(x,y)=(x+6)(y+7)(3x+2y)`

Solution

Find the maximum value of the following function: 

`f(x,y)=(x+6)(y+7)(3x+2y)`

To do so, we need to use AM-GM inequation: 

                                    `x+y+z` ≥`3.(xyz)^(1/3)`

                                    `\frac{4x+3y+13}3`≥`f(x,y)^(1/3)`

Then, we can see that :    `f(x,y)` ≤`((4x++3y+13)/3)^3`           

But, we knew that: `4x+3y=11`

Therefore: `f(x,y)`≤`((11+13)/3)^3=512`

Hence:  the maximum value of the following function: 

`f(x,y)=(x+6)(y+7)(3x+2y)` is `512`

Solution by: Thin Sokkean