IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Tanya_verma

Let x, y and z be three positive numbers such that x+y+z = 12

then the maximum value of x²y³z is.....

1.2676

2.5692

3.6912

4 7836

Do give the soln..

ashish_banga

i am getting 864
please check the question again

ramkumar_november

the answer is 6912 here is the solution::

given that x+y+z=12 .... it can be rearranged as ...

$2\bigg(\frac{x}{2}\bigg)+3\bigg(\frac{y}{3}\bigg)+z=12$ .........(1)

now we use weighted AM>=GM that is

$\frac{m_1a_1+m_2a_2+m_3a_3}{m_1+m_2+m_3}\geq (a_1^{m_1}.a_2^{m_2}.a_3^{m_3})^{\frac{1}{m_1+m_2+m_3}}$

put $m_1=2\;\;\;m_2=3\;\;m_3=1$

and $a_1=\frac{x}{2}\;\;a_2=\frac{y}{3}\;\;a_3=z$

$\frac{2\bigg(\frac{x}{2}\bigg)+3\bigg(\frac{y}{3}\bigg)+z}{2+3+1}\geq\Bigg(\bigg(\frac{x}{2}\bigg)^2.\bigg(\frac{y}{3}\bigg)^3.z\Bigg)^{\frac{1}{6}}$

$\frac{12}{6}\geq\Bigg(\frac{x^2.y^3.z}{108}\Bigg)^{\frac{1}{6}}$ from(1)

$x^2.y^3.z\leq 108 . 2^6$

$x^2.y^3.z\leq 6912$

so max value is 6912 ..... option 3 is correct....

ashish_banga

oh sorry
i thought it is cube root but it was sixth root
ramkumar is correct

anchitsaini

spideyunlimited

x+y+z = 12

x/2 + x/2 + y/3 + y/3 + y/3 + z = 12

Use AM > GM inequality
(x + y + z) / 6 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2 > [ (x^2 / 4)(y^3 / 27)(z) ]^1/6
2^6 > (x^2 / 4)(y^3 / 27)(z)
2^6 * 4*27 > x^2.y^3.z
x^2.y^3.z < 6912

so option 3

spideyunlimited

shit i should have refreshed the page,
had left it open since a long time.!

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