Find the maximum value of given expression $$ x+2y+3z = 15 $$
Find the maximum value of
$$ 6(1+x)yz + x(2y+3z) $$
I substituted $2y+3z=15-x$
And tried to use AM GM to find maximum value of $xyz$ but they don't occur at same value. I know that this is not the right way to do it but I don't have any idea on how to do these type of questions.
How do I proceed?
Edit
Do note that $x,y,z$ are positive real numbers.
 A: Let $x=5a$, $y=\frac{5}{2}b$ and $z=\frac{5}{3}c$.
Thus, $a+b+c=3$ and  since $(a+b+c)^2\geq3(ab+ac+bc),$ by AM-GM we obtain:
$$ 6(1+x)yz + x(2y+3z)=25(5abc+ab+ac+bc)\leq$$
$$\leq25\left(5\left(\frac{a+b+c}{3}\right)^3+\frac{1}{3}(a+b+c)^2\right)=200.$$
The equality occurs for $a=b=c=1$, which says that we got a maximal value.
A: Variational Approach
To maximize
$$
6(1+x)yz+x(2y+3z)\tag1
$$
under the constraint
$$
x+2y+3z=15\tag2
$$
we need
$$
(6yz+2y+3z)\,\delta x+(6z+6xz+2x)\,\delta y+(6y+6xy+3x)\,\delta z=0\tag3
$$
for all $\delta x$, $\delta y$, and $\delta z$ so that
$$
\delta x+2\delta y+3\delta z=0\tag4
$$
This means there is a $\lambda$ so that
$$
\begin{align}
\lambda&=6yz+2y+3z\implies(3z+1)(2y+1)=\lambda+1\\
2\lambda&=6z+6xz+2x\implies(3z+1)(x+1)=\lambda+1\\
3\lambda&=6y+6xy+3x\implies(2y+1)(x+1)=\lambda+1
\end{align}\tag5
$$
which has solutions
$$
(x,y,z)=\left(6,3,2\right)\mu\tag6
$$
Letting $\mu=\frac56$, gives
$$
(x,y,z)=\left(5,\frac52,\frac53\right)\tag7
$$
which satisfies $(2)$ and gives
$$
6(1+x)yz+x(2y+3z)=200\tag8
$$

Another Approach
Maximize
$$
(x+1)(2y+1)(3z+1)-(x+1)-(2y+1)-(3z+1)+2\tag9
$$
under the constraint
$$
(x+1)+(2y+1)+(3z+1)=18\tag{10}
$$
Considering the constraint, we are maximizing
$$
(x+1)(2y+1)(3z+1)-16\tag{11}
$$
under the constraint
$$
(x+1)+(2y+1)+(3z+1)=18\tag{12}
$$
This is the same as, under the constraint $a+b+c=18$, maximizing $abc=ab(18-a-b)$. Since
$$
0=\frac{\partial}{\partial a}ab(18-a-b)=b(18-2a-b)\tag{13}
$$
and
$$
0=\frac{\partial}{\partial b}ab(18-a-b)=a(18-a-2b)\tag{14}
$$
and since $a,b\ne0$, we have
$$
18-2a-b=0\tag{15}
$$
and
$$
18-a-2b=0\tag{16}
$$
Subtracting $(15)$ and $(16)$ gives that $a=b$, then back-substituting, we get
$$
a=b=c=6\tag{17}
$$
Applying $(17)$ to $(11)$ and $(12)$ gives $x+1=2y+1=3z+1=6$, which gives the maximum
$$
6^3-16=200\tag{18}
$$
A: With $x=15-2y-3z$ you need to find the maximum of
$$6(16-2y-3z)yz + (15-2y-3z)(2y+3z).$$
Then cancelling the partial derivatives,
$$\begin{cases}30+84z-8y-24yz-18z^2=0,\\45−18 +84 −36  −12 ^2=0.\end{cases}$$
Now if we center the first quadric,
$$30+84z-8y-24yz-18z^2=-18\left(z+\frac13\right)^2-24(y-4)\left(z+\frac13\right)=\left(z+\frac13\right)(90-18z-24y),$$ we can reject $z=-\dfrac13$.
If we center the second quadric,
$$45−18 +84 −36  −12 ^2=-36\left(y+\frac12\right)\left(z-\frac83\right)-12\left(y+\frac12\right)^2=\left(y+\frac12\right)(90-36z-12y)$$ where we can reject $y=-\dfrac12$.
Then the solution of the remaining linear system is 
$$y=\frac52,z=\frac53$$ and
$$x=5.$$
A: Doing the same as Yves Daoust, using
$$30+84z-8y-24yz-18z^2=0\tag 1$$ $$45−18 +84 −36  −12 ^2=0\tag 2$$ From $(1)$, extract $y=-\frac{3}{4} (z-5)$.
Replace in $(2)$ to get $$\frac{9}{4} \left(9 z^2-66 z+85\right)=0$$ the roots of the quadratic in $z$ are $\frac 53$ and $\frac {17}3$ to which correspond two values of $y$ but one must be discarded since non positive.
Just continue.
