ISI2004-MIII: 12

Question

The maximum possible value of $xy^2z^3$ subjected to condition $x,y,z \geq 0$ and $x+y+z=3$ is

• A. $1$
• B. $\frac{9}{8}$
• C. $\frac{9}{4}$
• D. $\frac{27}{16}$

Answer 1

Given ,

$x + y + z = 3$

$\Rightarrow  x +\left(\frac{y}{2}\right) +\left(\frac{y}{2}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) = 3.$

Now using A.M. G.M inequality , we have : 

$\dfrac{\left[x +\left(\frac{y}{2}\right) +\left(\frac{y}{2}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right) +\left(\frac{z}{3}\right)\right]}{6}$

 $\geq \large\left[x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right]^{\frac{1}{6}}$

$\Rightarrow  \large\left(x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right)^{\frac{1}{6}}\leq \frac{1}{2}$

$\Rightarrow  \large\left(x .\left(\frac{y}{2}\right) .\left(\frac{y}{2}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right) .\left(\frac{z}{3}\right)\right)\leq \frac{1}{64}$

$\Rightarrow x . y^{2} . z^{3}  \leq \frac {108}{64}  =\frac {27}{16}.$

Hence  D) is the correct answer

Comments

@Gate Fever

its straight forward AM of x,y,z = $\frac{x+y+z}{3}$

and GM of X,Y,Z is $\sqrt[3]{x+y+z}$

for purpose of this question we are making some changes in AM and GM and using the inequality 

---

satisfy the values 1/2,1,3/2 which is equal to 3

---

Right!

Answer 2

Using Langrage multiplier

$$F=xy^2z^3$$
$$\frac{\partial F}{\partial x}= y^2z^3 $$
$$\frac{\partial F}{\partial y}= 2xyz^3-$$
$$\frac{\partial F}{\partial z}= 3xy^2z^2$$
Now Equation of constraint is Q= $x+y+z = 3$
so,
$$\frac{\partial Q}{\partial x}= 3$$
$$\frac{\partial Q}{\partial y}= 3$$
$$\frac{\partial Q}{\partial z}= 3$$
Now what langrage mutiplier is suggesting is $F_x=\lambda Q_x$ like wise for $y\text{ and }z$
so,
$$\frac{\partial F}{\partial x}= y^2z^3=3\lambda---(i)$$
$$\frac{\partial F}{\partial y}= 2xyz^3=3\lambda---(ii)$$
$$\frac{\partial F}{\partial z}= 3xy^2z^2=3\lambda---(iii)$$
$$\frac{(ii)}{(i)}=\frac{2x}{y}=1 -----> 2x=y$$
$$\frac{(iii)}{(ii)}=\frac{3y}{2z}=1  -----> 3y=2z$$
$$6x=3y=2z$$
Putting in equation in Q we get
$$z=\frac{3}{2}\text{ and } y=1 \text{ and } x = \frac{1}{2}$$
Putting these value in F we get,
$$\frac{27}{16}$$
Anwser is D