A Maths問題
2008-03-29 7:29 am
1.N=x^2+y^2,where x and y are positive real numbers with xy=25.Find the values of x and y such that N is minimum.
回答 (3)
2008-03-29 7:44 am
2008-03-29 7:49 am
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
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
2008-03-29 7:45 am
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。
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。
收錄日期: 2021-04-23 19:56:39
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