The rate of transmission in a telegraph cable is observed to be proportional to x2ln(1/x) where x is the ratio of the radius of
1 answer:
Answer:
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.
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Answer:
When we have a quadratic equation:
a*x^2 + b*x + c = 0
There is something called the determinant, and this is:
D = b^2 - 4*a*c
If D < 0, then the we will have complex solutions.
In our case, we have
5*x^2 - 10*x + c = 0
Then the determinant is:
D = (-10)^2 - 4*5*c = 100 - 4*5*c
And we want this to be smaller than zero, then let's find the value of c such that the determinant is exactly zero:
D = 0 = 100 - 4*5*c
4*5*c = 100
20*c = 100
c = 100/20 = 5
As c is multiplicating the negative term in the equation, if c increases, then we will have that D < 0.
This means that c must be larger than 5 if we want to have complex solutions,
c > 5.
I can not represent this in your number line, but this would be represented with a white dot in the five, that extends infinitely to the right, something like the image below:
