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 Three numbers (Posted on 2003-11-17)
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have ?

 No Solution Yet Submitted by Ravi Raja Rating: 3.7500 (4 votes)

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 Solution by Lagrange Multipliers | Comment 18 of 21 |
Let f(x,y,z)=x, g(x,y,z)=x+y+z-5, h(x,y,z)=xy+yz+zx-3. Then subject to the constraints g=0 and h=0, f is stationary for any real x,y,z,L,M that satisfy g=h=0 and grad(f)=Lgrad(g)+Mgrad(h) (L and M are the Lagrange multipliers, see, e.g. the Dover book "Advanced Calculus of Several Variables" by C.H. Edwards, Jr.) The gradient equation evaluates to the 3 scalar equations 1=L +M(y+z), 0=L+M(x+z), 0=L+M(x+y) and the last two of these imply y=z. Writing y for z in g=0 gives x=5-2y and putting this into h=0 and also writing y for z then gives 3y^2-10y+3=0 so that y=1/3 or 3. Hence x=13/3,y=z=1/3 or x=-1,y=z=3. These correspond respectively to the maximum and the minimum attained by x on the intersection curve of g=0 and h=0.
 Posted by Richard on 2003-11-18 23:22:18

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