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2013-01-02T20:28:36+02:00

(1-2).(2-3):(3-1) sonuç çıkmaz kardeşim

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2013-01-02T20:29:04+02:00

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(1) x + y + z = 5
(2) xy + yz + zx = 3

I do not see how this can be solved with elementary algebra unless we use some insights from geometry. 
Namely, equation (1) is an equation of a plane in 3D space. 
When two equations are combined through (1)2 - 2(2), we get:

(3) x2 + y2 + z2 = 19. 

This equation is equation of the sphere whose radius is √19.
Radius is less than 5, which means that the intersection of the plane (1) and the sphere (3) is a circle. 
The intersecting circle is symmetric with respect to the planes x = y, y = z and x = z. 
(All these planes will cut the circle in two halves through its diameter).
So, we can "see" that both, minimum and maximum value of x on the intersecting circle will occur when y = z.
Using this insight we can then substitute y = z into both of the original equations to get:

(1) x + 2y = 5
(2) xy + y2 + xy = 3


This can be solved for x and y.

(1) x = 5 - 2y
(2) y2 + 2xy = 3

By substituting x from (1) in (2), we will end up with:

y2 + 2(5 - 2y)y = 3, which simplifies to:
3y2 - 10y + 3 = 0

Using quadratic formula we can get the values for y to be:
y1 = 1/3, and y2 = 3.

The first value of y, y1, yields x to be:
x = 5 - 2(1/3) = 13/3

The second value of y, y2, yields x to be:
x = 5 - 2(3) = -1.

Hence, the minimum value of x is -1 and the maximum value of x is 13/3.

Normally, this type of problem can be solved as a constrained optimization problem, which uses Lagrange multipliers and calculus to get the extreme points. I am not sure if there is "an easy" algebraic solution to the problem.
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