Consider the set S of points $(x,y)\in R^2$ which minimize the real valued function
$f(x,y)=(x+y-1{)}^2+(x+y{)}^2$
Which of the following statements is true about the set S?

**Minimizing the Function $f(x,y)$ **

We are asked to find the set S of points $(x,y)$ that minimize the function $f(x,y)=(x+y-1{)}^2+(x+y{)}^2$. Let $u=x+y$. Substituting this into the function, we get a function of a single variable $u$: $f(u)=(u-1{)}^2+u^2$

Expand the expression: $f(u)=(u^2-2u+1)+u^2$ $f(u)=2u^2-2u+1$

Finding the Minimum Value

The function $f(u)=2u^2-2u+1$ is a quadratic function, representing an upward-opening parabola. Its minimum occurs at the vertex. We can find the value of $u$ at the minimum by taking the derivative with respect to $u$ and setting it to zero.

The derivative is: $\frac{df}{du}=\frac{d}{du}(2u^2-2u+1)=4u-2$

Set the derivative to zero to find the critical point: $4u-2=0$ $4u=2$ $u=\frac{1}{2}$

The minimum value of the function $f(u)$ occurs when $u=1/2$. The minimum value is $f(1/2)=(1/2-1{)}^2+(1/2{)}^2=(-1/2{)}^2+(1/2{)}^2=1/4+1/4=1/2$.

Defining the Set S

The set S consists of all points $(x,y)$ for which $f(x,y)$ attains its minimum value. This happens when $u=x+y=1/2$. So, the set S is the set of all points $(x,y)$ such that:

$x+y=\frac{1}{2}$

Characterizing the Set S

The equation $x+y=1/2$ defines a straight line in the $R^2$ plane. This line represents all possible pairs of $(x,y)$ that satisfy the condition for the minimum value of $f(x,y)$.

A straight line contains infinitely many points. Therefore, the set S is infinite.