Question

if x,y,z>0 such that x+2y+3z=15,then the maximum value of 6(1+x)yz+x(2y+3z) is

Answer 1

Step-by-step explanation:

SOLUTION:

Put X+2y+3z=15 as equation 1 and 2x+y + z=2 as equation 2

Multiply equation 2 with 3 so it becomes

6x+ 3y+3z= 6 and substract equation 1 from it

6x+3y+3z=6

-x-2y-3z=-15 (minus sign is multiplied to the whole equation)

3z gets eliminated and we are left with 5x+y=-9

Now substract x+y = 7 from this

5x+y=-9

-x-y=-7 (minus multiplies to whole equation)

y gets eliminated We get 4x=-16 and so x=-4

Since x +y= 7 we can write -4+y=7 ans so y =11

Now put the values of x and y in any of the two equations. Suppose I put them in equation 1

-4 +2(11)+3z= 15

Solving this

-4+22+3z= 15

18+3 z=15

3z= 15–18

3z=-3

z=-1

Then x+ y + z = -4+11–1=6