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The maximum value of the function

$f\left(x, y, z\right)= \left(x - 1 / 3\right)^{2}+ \left(y - 1 / 3\right)^{2}+ \left(z - 1 / 3\right)^{2}$

subject to the constraints

$x + y + z=1,\quad x \geq 0, y \geq 0, z \geq 0$

is

  1. $1 / 3$
  2. $2 / 3$
  3. $1$
  4. $4 / 3$
  5. $4 / 9$
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2 Answers

Best answer
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7 votes
Expanding given equation

$x^2 + y^2 + z^2 - \frac{2}{3}(x+y+z) + \frac{1}{3}$

$\quad = x^2 + y^2 + z^2 - \frac{2}{3} + \frac{1}{3}$

$\quad = (x+y+z)^2 - \frac{1}{3} - 2(xy + yz +xz)$

$\quad = 1 - \frac{1}{3} - 2(xy + yz +xz)$

$\quad = \frac{2}{3} - 2(xy + yz +xz)$

Now to maximize it, we need to minimize $(xy + yz + xz)$. As all $x,y$ and $z$ are non-negative $ xy + yz + xz \geq 0.$ So, the maximum value is $\frac{2}{3}$.

Correct Option: B.
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Since all X,Y,Z are greater than equal to zero. So in X+Y+Z=1 means any two out of three variable will be Zero and remaining variable will have value one.. 

So, Max value will be 

(1-1/3)2+(0-1/3)2+(0-1/3)2=2/3

Answer:

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