Find an equation of the plane that passes through the point (1,2,3) and cuts off the smallest volumn in the first octant?Don't use lagrange multipliers.
Use second derivatives test to find
find an eq of the plane
2012-05-02 4:46 am
回答 (2)
2012-05-03 11:29 am
✔ 最佳答案
http://s1152.photobucket.com/albums/p489/minglamchoi/PlainCuts1stOctantWithSmallestVolume.jpg圖片參考：http://s1152.photobucket.com/albums/p489/minglamchoi/PlainCuts1stOctantWithSmallestVolume.jpg
The 2nd derivative is kind of simple and is not included here.
2012-05-03 04:14:40 補充：
The last part solving a and b is wrong. The right answer should be
2b - a = 0 so a = 2b, putting back to (A) we have
2b^2 - 4b^2 - 3b = -2b^2 - 3b = b(-2b -3) = 0
so b = 0 or b = -3/2
so a = 2b = -3 so the equation of the plain is
z - 3 = -3(x-1) -3(y-2)/2
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