TIFR CSE 2012 | Part A | Question: 13

Question

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

• A. $1 / 3$
• B. $2 / 3$
• C. $1$
• D. $4 / 3$
• E. $4 / 9$

Answer 1

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.

Comments

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.

---

I solved it and my first derivative is

$x + y + z = 1$

If I put the values of any two variables as zero and the third as one. How can I assure the function value is maximum?

Answer 2

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

Comments

Ans is 4/9.

---

How,?

I'm getting 16/9 after solving the above equation.

---

@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.

---

"So in X+Y+Z=1 means any two out of three variable will be Zero and remaining variable will have value one."  why?

we can have x=0.5, y=0.5 and z=0 ,