(2013•温州一模)若变量x,y满足不等式x−y−1≥0y≥1
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(2013•温州一模)若变量x,y满足不等式
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![(2013•温州一模)若变量x,y满足不等式x−y−1≥0y≥1](/uploads/image/z/16731510-6-0.jpg?t=%EF%BC%882013%E2%80%A2%E6%B8%A9%E5%B7%9E%E4%B8%80%E6%A8%A1%EF%BC%89%E8%8B%A5%E5%8F%98%E9%87%8Fx%EF%BC%8Cy%E6%BB%A1%E8%B6%B3%E4%B8%8D%E7%AD%89%E5%BC%8Fx%E2%88%92y%E2%88%921%E2%89%A50y%E2%89%A51)
![](http://img.wesiedu.com/upload/d/14/d140c03bc885705dcc57f58f1b2de5ee.jpg)
z=x2+y2表示(0,0)到可行域的距离的平方,
当原点到点A(2,1)时,距离最小,
则y2+x2的最小值是(0,0)到(2,1)的距离的平方:5,
则z=x2+y2的最小值是5.
故答案为:5.
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