The value of x that gives the maximum transmission is 1/√e ≅0.607
Step-by-step explanation:
Lets call f the rate function f. Note that f(x) = k * x^2ln(1/x), where k is a positive constant (this is because f is proportional to the other expression). In order to compute the maximum of f in (0,1), we derivate f, using the product rule.
We need to equalize f' to 0
k*(2x ln(1/x) - x) = 0 -------- We send k dividing to the other side
2x ln(1/x) - x = 0 -------- Now we take the x and move it to the other side
2x ln(1/x) = x -- Now, we send 2x dividing (note that x>0, so we can divide)
ln(1/x) = x/2x = 1/2 ------- we send the natural logarithm as exp
1/x = e^(1/2)
x = 1/e^(1/2) = 1/√e ≅ 0.607
Thus, the value of x that gives the maximum transmission is 1/√e.
This would be a guess but I think you would need two numbers in the hundreds that also have 2 parts of which have a product of a number ending in 0, like 4 and 5.
