Respuesta :
Answer:
x = 0.606
Step-by-step explanation:
Data provided in the question:
rate of transmission is proportional to [tex]x^{2}ln(\frac{1}{x})[/tex]
or
rate of transmission, R = [tex]C[x^{2}ln(\frac{1}{x})][/tex]
here, C is the proportionality constant
Now,
for point of maxima
differentiating the function with respect to 'x'
R' = [tex]\frac{d(C[x^{2}ln(\frac{1}{x})])}{dx}[/tex]
using the product rule, we get\
R' = [tex]C[-2x^{1}\ln\left(x\right)-x^{1}][/tex]
or
R' = [tex]C[-x^{1}\left(2\ln\left(x\right)+1\right)[/tex]
Now,
R' = 0 [for point of maxima]
OR
[tex]C[-2x^{1}\ln\left(x\right)-x^{1}][/tex] = 0
or
-x²¹ = 0 or [tex]\left(2\ln\left(x\right)+1\right)[/tex] = 0
or
ln(x) = [tex]\frac{-1}{2}[/tex]
or
x = 0.606
Since,
0 < x < 1
Hence,
accepted value of x = 0.606
