Initial feasible solution for barrier method

Question

From this example page 9 It said set initial feasible solution at 2
here's barrier function: $$T(x)=\frac{100}{x}+\frac{1}{r}(\frac{-1}{x-5})$$ after derivative: $$\frac{\delta T}{\delta x}=\frac{1}{r(x-5)^2}-\frac{100}{x^2}$$ we show x in terms of r (we skip the step of algebra): $$x=\frac{-50r}{1-10r}$$ then we set r to infinite we get x =5
x never inolve the iteration
Why we need an initial value
and form my refrence book it said barrier function like that: $$T(x)=\frac{100}{x}-r(\frac{1}{x-5})$$ but the result gives -5 I think because it is absent of lower bound
which one should we choose for barrier method

Answer 1

Simplifying your equation we obtain $x^2(1-100r)+100rx-2500r=0$ using the quadratic formula we get $$x_1=\frac{50 \left(10 r-\sqrt{r}\right)}{100 r-1}$$ $$x_2=\frac{50 \left(10 r+\sqrt{r}\right)}{100 r-1}$$