# 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.

• are $x,y,z$ real numbers? – Dr. Sonnhard Graubner Feb 21 '18 at 9:22
• there is no global maximum – Dr. Sonnhard Graubner Feb 21 '18 at 9:25
• @Dr.SonnhardGraubner yes sir – user481779 Feb 21 '18 at 9:25
• why is this question downvoted? The OP has stated what he tried. – Rohan Shinde Feb 21 '18 at 9:38

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.

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 $$6^3-16=200\tag{18}$$

• Your approach is very interesting. We can also use AM GM in 11 – user481779 Feb 21 '18 at 20:39
• Indeed, but that approach was used by Michael Rozenberg, so I took a different path. – robjohn Feb 21 '18 at 21:37

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,

$$\begin{cases}30+84z-8y-24yz-18z^2=0,\\45−18 𝑧+84 𝑦−36 𝑦 𝑧−12 𝑦^2=0.\end{cases}$$

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,

$$45−18 𝑧+84 𝑦−36 𝑦 𝑧−12 𝑦^2=-36\left(y+\frac12\right)\left(z-\frac83\right)-12\left(y+\frac12\right)^2=\left(y+\frac12\right)(90-36z-12y)$$ where we can reject $y=-\dfrac12$.

Then the solution of the remaining linear system is

$$y=\frac52,z=\frac53$$ and

$$x=5.$$

Doing the same as Yves Daoust, using $$30+84z-8y-24yz-18z^2=0\tag 1$$ $$45−18 𝑧+84 𝑦−36 𝑦 𝑧−12 𝑦^2=0\tag 2$$ 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.