Maths-matrix

2011-05-08 4:50 pm
x-2y+z=5Consider the systemn of linear equations(*){x-y-z=4 , where a is a real number.x+y+az=2a) Find the range of values of a such that(*) ha s a unique solution. Solve (*) when it has a unique solution.(I find that a≠5;x=3, y=-1, z=0) b) Suppose that a=-5. Solve (*)(I solve that x=t, y=(2t-9)/3, z=(t-3)/3x-2y+z=5c) Supper that (x,y,z)satisfies{x-y-z=4 . Find the least values of (x^2)+(9y^2)+3z and x+y+az=2the corresponding valuesof x,y,z. a, b part 我已經計了,麻煩幫忙計c part
更新1:

x-2y+z=5 (*){x-y-z=4 x+y+az=2

更新2:

點解要將佢=0?

更新3:

2t + 4(2t - 9) + 1<<---起邊度計出黎?

回答 (2)

2011-05-08 6:06 pm
✔ 最佳答案
(c) x = t, y = (2t - 9)/3, z = (t - 3)/3

(x^2) + (9y^2) + 3z

= t^2 + (2t - 9)^2 + (t - 3)

differentiate w.r.t t and set it equal to 0

2t + 4(2t - 9) + 1 = 0

10t = 35

t = 3.5

So, x = 3.5, y = -2/3, z = 1/6

Least value = 16.75

2011-05-08 18:16:30 補充:
用微分法求最小值
2011-05-08 9:05 pm
2t + 4(2t - 9) + 1<<---起邊度計出黎?


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