Min

Jan 2013
96
0
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!
 
Mar 2013
90
0
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 \(\displaystyle +\infty\).
 
Jan 2013
96
0
Really? I sugest you to read again the statement!
 
Jun 2013
1,315
116
London, England
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
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
\(\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)\)