# Min

#### yo79

Find the minimum of $$\displaystyle x^2-2x+y$$ if $$\displaystyle x,y \ge 0$$ and $$\displaystyle x+y \le 2$$. I need a graphic interpretation of the problem (in the plane xOy). Thank you!

#### Nehushtan

There is no minimum. For example, if you fix $$\displaystyle x=0$$, then $$\displaystyle k=x^2-2x+y=y$$ and then, since you only need $$\displaystyle y\leq2$$, you can take $$\displaystyle y$$, and hence $$\displaystyle k$$, arbitrarily close to $$\displaystyle -\infty$$.

There is no maximum either. For example, if you fix $$\displaystyle y=0$$, then $$\displaystyle k=x^2-2x+y=x^2-2x$$ and then, since you only need $$\displaystyle x\leq2$$, taking $$\displaystyle x$$ arbitrarily close to $$\displaystyle -\infty$$ will make $$\displaystyle k$$ arbitrarily close to $$\displaystyle +\infty$$.

#### yo79

Really? I sugest you to read again the statement!

#### Pero

The min must be at (1, 0). For every value of y, x^2 - 2x will have the same minimum value of -1 (at x = 1). So, the minimum will be -1 at (1, 0).

#### greg1313

Forum Staff
$$\displaystyle f(x,\,y)\,=\,x^2\,-\,2x\,+\,y\,=\,(x\,-\,1)^2\,+\,y\,-\,1\,\Rightarrow\,\min\(f(x,\,y)$$\,=\,-1\text{ at }(x,\,y)\,=\,(1,\,0)\)