268 views

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 | 268 views

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 $\frac{2}{3}$.
selected
0
Hello dhruv

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

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

+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.

1
2
+1 vote