题目内容
变量x、y满足
(1)假设z =4x-3y,求z的最大值.
(2)设z =
,求z的最小值.
(3)设z =x2+y2,求z的取值范围.
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(1)假设z =4x-3y,求z的最大值.
(2)设z =
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(3)设z =x2+y2,求z的取值范围.
(1)zmax=14;(2)zmax=koB=
;(3)z
.
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(1)作出不等式表示的可行域,然后根据直线z=4x-3y与在y轴的截距是正相关,还是负相关来确定最优解,从而求得最大值.
(2)注意利用其几何意义 z =
,它表示可行域内的点与原点连线的斜率的大小.注意数形结合即可求解.
(3) z =x2+y2它的几何意义是可行域内的点到原点距离的平方,利用它然后数形结合即可.
解:联立
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4’
(1)zmax=14 6’
(2)zmax=koB=
9’
(3)z
13’
(2)注意利用其几何意义 z =
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(3) z =x2+y2它的几何意义是可行域内的点到原点距离的平方,利用它然后数形结合即可.
解:联立
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(1)zmax=14 6’
(2)zmax=koB=
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(3)z
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