A Maths問題

Please refer to the following solution:

圖片參考：http://hk.geocities.com/stevieg_1023/XYAMA.gif

N=x^2＋y^2
N=x^2+(25/x)^2
N=x^2+625/(x^2)
dN/dx=2x-1250/(x^3)
When dN/dx=0,
2x-1250/(x^3)=0
2x^5-1250x=0
x(2x^4-1250)=0
x=0 or 2x^4=1250
x=0 or x=5 or x=-5
d^2N/dx^2=2＋3750/x^4
when x=0,d^2N/dx^2 is undefined
when x=5,-5,d^2N/dx^2=6＞0
N is minimum when x=5 and x=-5
When x=5,y=5
WhenX=-5,y=-5

2008-03-28 23:50:17 補充：
Sorry, x is positive

應用微分法。
思路：將N=x^2+y^2變成只得兩個變數的方程
Since xy=25
N=x^2+(25/x)^2

dN/dx = 2x - 2(625)/(x^3)

when dN/dx=0 , x=5 or -5(rejected, since x is positive)

differentiate once more, we get a positive number
hence when x=5, N is minimum

therefore, when x=y=5, N=50 is minimum.

2008-03-28 23:46:32 補充：
注意：x is positive, 不會是-5。