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+2 votes
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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

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

is

  1. $1 / 3$
  2. $2 / 3$
  3. $1$
  4. $4 / 3$
  5. $4 / 9$
asked in Calculus by Boss (41.2k points) | 268 views

2 Answers

+4 votes
Best answer
I think the answer is 2/3. Verification required.

Expanding given equation

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

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

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

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

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

Now to maximize it, we need to minimize $(xy + yz + xz)$. As all x,y and z are non-negative xy + yz + xz >= 0. so the maximum value is $\frac{2}{3}$.
answered by Active (2k points)
selected by
0
Hello dhruv

You answer and way to solve it , is absolutely correct.

Please don't start your answer like 'i think'. when you're not sure about answer , prefer to answer in form of comment.
+2 votes

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

answered by Active (2.3k points)
+2
Ans is 4/9.
0
How,?

I'm getting 16/9 after solving the above equation.
0
@Shivam Dwivedi

I don't think so, did you consider the constraint x + y + z = 1? If you did, can you please tell how you got 4/9.

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