Michael's maximizing mania

By the arithmetic-geometric mean inequality, $\frac{2x+y+y+y+y+6z}{6}\ge(2xy^46z)^{\frac{1}{6}}$ Plugging in the expression $x+2y+3z=12$ , we get $\frac{24}{6}=4\ge(2xy^46z)^{\frac{1}{6}}$ Raising each side to the $6^{th}$ power yields $4^6=2^{12}\ge2xy^46z$ or $\frac{2^{10}}{3}\ge xy^4z$ Thus $\frac{3}{2}xy^4z\le\frac{3}{2}\cdot\frac{2^{10}}{3}=2^9=512$ However, we must check that this value can be obtained. In the AM-GM inequality, equality is obtained if each term is equal. Thus we have $2x=y=6z$ . Plugging this into the original equation gives $\frac{y}{2}+2y+\frac{y}{2}=12\Rightarrow y=4\Rightarrow(x,y,z)=(2,4,\frac{2}{3})$ Now we have $\frac{3}{2}xy^4z=\frac{3}{2}(2)(4)^4(\frac{2}{3})=2^9$ So this value can be obtained, and $\boxed{512}$ is our answer.

This problem is completely observation based.

We need to apply AM-GM inequality but not in a direct manner. We need to do some mathematical jugglery with $x+2y+3z$ so as to get the maximum value of $\frac{3}{2}xy^4z$

Observe that, we can write $x+2y+3z = x+ \frac{y}{2}+ \frac{y}{2}+ \frac{y}{2}+ \frac{y}{2} + 3z = 12$

Now, we can apply AM-GM inequality to easily get our answer

$\frac{x+ \frac{y}{2}+ \frac{y}{2}+ \frac{y}{2}+ \frac{y}{2} + 3z}{6} \geq (\frac{3}{16}xy^4z)^{\frac{1}{6}}$

$(\frac{3}{16}xy^4z)^{\frac{1}{6}} \leq \frac{12}{6} = 2$

$\frac{3}{16}xy^4z \leq 2^6 = 64$

$\frac{3}{2}xy^4z \leq 64\times8 = 512$

Hence, the maximum value of $\frac{3}{2}xy^4z$ is $\boxed{512}$

Since we see $y^4$ so we will manipulate $2y$ as $\displaystyle\frac{y}{2}$ $\displaystyle+\frac{y}{2}+$ $\displaystyle\frac{y}{2}+$ $\displaystyle\frac{y}{2}$ . Now employing AM-GM inequality as the numbers are positive, we have

$\displaystyle\frac{x+\frac{y}{2}+\frac{y}{2}+\frac{y}{2}+\frac{y}{2}+3z}{6}\geq\left(\frac{3xy^4z}{16}\right)^{\frac{1}{6}}$

Substituting $\displaystyle\,x+2y+3z$ with $12$ and raising both sides to $\displaystyle\,6^{th}$ power we have $\frac{3}{2}xy^4z\leq{512}$ .

We know $x + 2y + 3z = 12$ so $x + \dfrac12 y + \dfrac12 y + \dfrac12 y + \dfrac12 y + 3z = 12$ . From AM-GM, we know

$2 \ge \dfrac{x + \dfrac12 y + \dfrac12 y + \dfrac12 y + \dfrac12 y + 3z}{6} \ge \sqrt[6]{x \cdot \left(\dfrac12 y\right)^4 \cdot 3z} \ge \sqrt[6]{\dfrac{3x^4z}{16}}$ Therefore $64 \ge \dfrac{3x^4z}{16}$ so $512 = 64 \cdot 8 \ge \dfrac32 x^4z$ . Equality occurs when $x = 2, \ y=4, \ z=\dfrac23$ so our answer is $\boxed{512}$ .

Notice that we can "break" up the given equation into

$x + \frac{y}{2} + \frac{y}{2} + \frac{y}{2} + \frac{y}{2} + 3z = 12$

But by AM-GM, we have

$x + \frac{y}{2} + \frac{y}{2} + \frac{y}{2} + \frac{y}{2} + 3z \geq \sqrt[6]{x\frac{y^4}{16}3z}$

$12 \geq \sqrt[6]{x\frac{y^4}{16}3z}$

which is equivalent to

$\frac{1024}{3} \geq xy^{4}z$

multiplying by $\frac{3}{2}$ yields

$512 \geq \frac{3}{2}xy^{4}z$

this means that the maximum value of $\frac{3}{2}xy^{4}z$ is $512$ .

$$ It took more than an hour for me to solve this problem because I have to learn the method of Lagrange multipliers first. I've never heard a such thing as Lagrange multipliers before. Nice problem since I get a new kind of knowledge.

Using the method of Lagrange multipliers :

Note that the values of $(x, z, y)$ that give the maximum of $\frac{3}{2}xy^4z$ also give the maximum of $xy^4z$ and of $\ln(xy^4z) = \ln(x) +4 \ln(y)+ \ln(z)$ since $x,y,z$ are all positive.

We set up the Lagrangian

$\mathcal{L} = \ln(x) +4 \ln(y)+ \ln(z) - \lambda(x+2y+3z-12)$

Now take all partial derivative and set to zero.

$\dfrac{\partial \mathcal{L}}{\partial x} = \frac{1}{x} - \lambda =0 \iff x = \frac{1}{\lambda}$

$\dfrac{\partial \mathcal{L}}{\partial y} = \frac{4}{y} - 2\lambda =0 \iff 2y = \frac{4}{\lambda}$

$\dfrac{\partial \mathcal{L}}{\partial z} = \frac{1}{z} - 3\lambda =0 \iff 3z = \frac{1}{\lambda}$

$\dfrac{\partial \mathcal{L}}{\partial \lambda} = -x-2y-3z+12 =0$

Plug into the last partial derivative to get value of $\lambda$ and we find $(x, z, y) = ( 2,4,2/3)$ .

Evaluating $\frac{3}{2}xy^4z$ at those values gives $\boxed{512}$ .

As x y z are positive reals we can apply AM-GM inequality. x +4 (y/2)+3z>(3/16 xy^4z)^1/6 On solving we get the required value <= 512. Hence its maximum value is 512.

Whenever I see questions like this, I start out by trying the AM-GM inequality. Since we want a product of $x$ , $y^{4}$ and $z$ on the Geometric Mean side, we rewrite the given equation to include 4 terms involving $y$ .

$x + 2y + 3z = x + \dfrac{y}{2} + \dfrac{y}{2} + \dfrac{y}{2} + \dfrac{y}{2} + 3z = 12$

Now, taking AM-GM, we get:

$\dfrac{x + \dfrac{y}{2} + \dfrac{y}{2} + \dfrac{y}{2} + \dfrac{y}{2} + 3z}{6} \ge \left(x*\dfrac{y}{2}*\dfrac{y}{2}*\dfrac{y}{2}*\dfrac{y}{2}*3z\right)^{\frac{1}{6}}$

$\dfrac{12}{6} \ge \left(\dfrac{3xy^{4}z}{16}\right)^{\frac{1}{6}}$

$2^{6} \ge \dfrac{3xy^{4}z}{16}$

$8 \times 2^{6} \ge 8 \times \left(\dfrac{3xy^{4}z}{16}\right)$

$512 \ge \dfrac{3}{2}xy^{4}z$

Hence, the maximum value of $\dfrac{3}{2}xy^{4}z$ is $\boxed{512}$

The first linear equation is equal to $x+\frac{1}{2}y+\frac{1}{2}y+\frac{1}{2}y+\frac{1}{2}y+3z=12.$ By AM-GM inequality, $\frac{x+\frac{1}{2}y+\frac{1}{2}y+\frac{1}{2}y+\frac{1}{2}y+3z}{6}$ is greater than or equal to $^{6}\sqrt{\frac{3}{16}xy^{4}z}$ (the product of the six terms on the numerator on the left hand side.) From this equation it can be seen that the largest possible value is $8 \times 2^{6}=512$ which can be achieved when $x=\frac{y}{2}=3z$ $(x=2, y=4, z=\frac{2}{3})$

A.M . > G.M

if positive real nos are given ,then atleast one times we should think about the A.M>=G.M,let x,y/2,y/2,y/2,y/2,3z , there are six positive no and after the multiplication of this nos we are getting xy^4z factor.then here we can apply A.M>=G.M (x+y/2+y/2+y/2+y/2+3z)/6>=6th root of (x y/2 y/2 y/2 y/2*3z).After solving this we can get the ANS.And the ans is 512.

Y cannot be 5 as the denominator is 2. therefore y has to be 4.taking z to be divisible by 3, z=1/3=.>x=3 and the product is 192. let z=2/3 and x=2 and the product is 512. Lagrangian approach is methodical but I took the intuitive route.

$x+2y+z$ can be written as

$x + (1/2)y + (1/2)y + (1/2)y + (1/2)y + 3z$

By AM - GM Inequality, we have

$x + (1/2)y + (1/2)y + (1/2)y + (1/2)y + 3z \ge 6\sqrt[6]{\frac{3}{16}xy^{4}z}$

But $x+2y+z = x + (1/2)y + (1/2)y + (1/2)y + (1/2)y + 3z = 12$ ; hence

$6\sqrt[6]{\frac{3}{16}xy^{4}z}\le12$

or

$\sqrt[6]{\frac{3}{16}xy^{4}z}\le2$

It follows that

$\frac{3}{2}xy^{4}z\le2^{9}$

Which means that the maximum value of $\frac{3}{2}xy^{4}z$ must be $2^{9}$ or $\boxed{512}$

Split 2y into y/2,+y/2+y/2+y/2....now apply am>gm to 6 quantities which also include x and 3z

first of all our focus goes to the "Largest possible value", hence we think of the AM>=GM concept. Now, to get the desired 3/2 x y^4 z we break 2y as y/2 + y/2 + y/2 + y/2 after which we will apply the AM>= GM for x, the broken y's and 3z therefore, we have $\frac{x + y/2 +y/2+y/2+y/2 +3z}{6}$ = (3/16 x y^4 z)^1/6) we know that the LHS =12. Hence on solving we get 3/2 x $y^{4}$ z <=512 Hence the answer $\boxed{512}$

x+2y+3z=12= (x+y/2+y/2+y/2+y/2+3z)..... <1> using AM>OR=GM in <1>, (x+y/2+y/2+y/2+y/2+3z)/6 >or= (x* y^4/16 3z)^1/6 2^6 >or= x (y^4)/16 3z 1024/3 >or= x y^4 z therefore, 3/2(x y^4 z) <or= (3/2) (1024/3) 3/2(x* y^4 *z) <or= 512