|February 21st, 2014, 10:45 AM||#2|
Joined: Mar 2013
There is no minimum. For example, if you fix , then and then, since you only need , you can take , and hence , arbitrarily close to .
There is no maximum either. For example, if you fix , then and then, since you only need , taking arbitrarily close to will make arbitrarily close to .
|February 26th, 2014, 03:16 AM||#4|
Joined: Jun 2013
From: 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).