設x,y屬於R試求（2x+2y+2） 平方+（x+3y+1)平方+(2x+4y_1)平方之最小值?
回答 (1)
2019-09-16 11:33 pm
Sol
F(x，y)=(2x+2y+2)^2+(x+3y+1)^2+(2x+4y+1)^2
∂F/∂x=2(2x+2y+2)*2+2(x+3u+1)+2(2x+y-1)*2
=(8x+8y+8)+(2x+6y+2)+(8x+16y-4)
=18x+30y+6
∂F/∂y=2(2x+2y+2)*2+2(x+3y+1)*3+2(2x+4y-1)*4
=(8x+8y+8)+(6x+18y+6)+(16x+32y-8)
=30x+58y+6
Set 18x+30y+6=0，30x+58y+6=0
3x+5y=－1，15x+29y=－3
5(3x+5y)－(15x+29y)=－5+3
y=1/2
3x+5/2=－1
3x=－7/2
x=－7/6
F(x，y)=4
F(x，y)=(2x+2y+2)^2+(x+3y+1)^2+(2x+4y+1)^2
∂F/∂x=2(2x+2y+2)*2+2(x+3u+1)+2(2x+y-1)*2
=(8x+8y+8)+(2x+6y+2)+(8x+16y-4)
=18x+30y+6
∂F/∂y=2(2x+2y+2)*2+2(x+3y+1)*3+2(2x+4y-1)*4
=(8x+8y+8)+(6x+18y+6)+(16x+32y-8)
=30x+58y+6
Set 18x+30y+6=0，30x+58y+6=0
3x+5y=－1，15x+29y=－3
5(3x+5y)－(15x+29y)=－5+3
y=1/2
3x+5/2=－1
3x=－7/2
x=－7/6
F(x，y)=4
收錄日期: 2021-04-30 22:49:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190916085837AACqj6K