三重积分题,
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三重积分题,
![](http://img.wesiedu.com/upload/5/c7/5c785078c4d5b79fde93aa20deeff136.jpg)
![](http://img.wesiedu.com/upload/5/c7/5c785078c4d5b79fde93aa20deeff136.jpg)
![三重积分题,](/uploads/image/z/17786065-49-5.jpg?t=%E4%B8%89%E9%87%8D%E7%A7%AF%E5%88%86%E9%A2%98%2C%26nbsp%3B)
方法一、采用偏微分方法
定义域D内的极大值点应满足:
z'(x) = 2xy(4-x-y) - x²y =0
z'(y) = x²(4-x-y) - x²y = 0
整理得:
xy(8-2x-3y) = 0;
y²(4-2x-y) = 0;
xy =0 ;与 y²=0为域D边界,暂不考虑;
解得:
x=2;y=1;
代入得 z = x²y(4-x-y) = 4
将域D的边界条件分别代入:
x=0:z = x²y(4-x-y) = 0;
y=0:z = x²y(4-x-y) = 0;
x+y = 6时 z = x²y(4-x-y) z ≤ (x³-4x²)²/(4x²) = (x²-4x)²/4 = [(x-2)²-4]²/4
显然 当 x=2 时,不等式右侧有最大值 4;
即 x=2时,z 有最大值4;
检验边界条件:
x=0:z = x²y(4-x-y) = 0;
y=0:z = x²y(4-x-y) = 0;
x+y = 6时 z = x²y(4-x-y)
定义域D内的极大值点应满足:
z'(x) = 2xy(4-x-y) - x²y =0
z'(y) = x²(4-x-y) - x²y = 0
整理得:
xy(8-2x-3y) = 0;
y²(4-2x-y) = 0;
xy =0 ;与 y²=0为域D边界,暂不考虑;
解得:
x=2;y=1;
代入得 z = x²y(4-x-y) = 4
将域D的边界条件分别代入:
x=0:z = x²y(4-x-y) = 0;
y=0:z = x²y(4-x-y) = 0;
x+y = 6时 z = x²y(4-x-y) z ≤ (x³-4x²)²/(4x²) = (x²-4x)²/4 = [(x-2)²-4]²/4
显然 当 x=2 时,不等式右侧有最大值 4;
即 x=2时,z 有最大值4;
检验边界条件:
x=0:z = x²y(4-x-y) = 0;
y=0:z = x²y(4-x-y) = 0;
x+y = 6时 z = x²y(4-x-y)