1.

Find the value of 1,x, yz1,y, xz 1z, xy​

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Answer:

1/X -1)(1/y-1)(1/z-1) = (1-x)(1-y)(1-z)/(xyz) = (1 -(x+y+z)+ (xy+yz+zx)-(xyz))/(xyz)

= { (xy+yz+zx)-(xyz) }/(xyz) = (1/x +1/y+1/z) -1.

Let ƒ(x, y, z) = 1/x +1/y+1/z -1.

Let us find the optimum ƒ(x, y, z) SUBJECT to (x+y+z)=1.

Let the objective function be ƒ(x, y, z) + µ(x+y+z-1)

∂ƒ/∂x = -1/(x^2)+µ = 0. ==> x^2 = 1/µ.

∂ƒ/∂y = 0.==> y^2 =1/µ.

∂ƒ/∂z =0. ==> z^2 = 1/µ.

x =y =z = ± 1/√µ.

∂ƒ/∂µ = 0 = (x+y+z-1).==> ±3/√µ -1. µ = ±√9.

(1/x+1/y+1/z) -1 = 3√µ -1 = 8. (Maximum) or -10 (MINIMUM)


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