at a single ticket booth, customers arrive randomly at a rate of x per hour.
the average line length is given by
F(x)= x^2/400-20x
Where 0<= x <20. To keep the wait in line reasonable, it is required that the average line length should not exceed 10 customers.
Determine the range of rates at which customers can arrive before a second attendant is needed. Express your answer in interval form.
10 = (1/400) x^2 - 20 x ??? you did not use parenthesis so I assume this
4000 = x^2 - 8000 x
x^2 - 8000 x - 4000 = 0
x = [ 8000 +/- sqrt(64^10^6+16000) ]/2
x =
nah, impossible you must mean
f(x) = x^2 /(400 - 20 x)
10 = x^2 / (400 - 20 x)
4000 -200 x = x^2
x^2 + 200 x - 4000 = 0
x = [ -200 +/- sqrt(40000+16000) ]/2
= [ -200 +/- 236 ]/2
= 18, -218 ignore -218
0 </= x </= 18
thank you bottom is the correct way.
To determine the range of rates at which customers can arrive before a second attendant is needed, we need to find the range of values for x that satisfy the condition F(x) ≤ 10.
Given:
F(x) = x^2/400 - 20x
0 ≤ x < 20
We need to find the values of x where F(x) ≤ 10.
F(x) ≤ 10
x^2/400 - 20x ≤ 10
To simplify the inequality, let's multiply all terms by 400 to get rid of the fraction:
400 * (x^2/400 - 20x) ≤ 400 * 10
x^2 - 8000x ≤ 4000
Rearranging the inequality:
x^2 - 8000x - 4000 ≤ 0
Now we can solve this inequality by finding the range of x:
The quadratic equation x^2 - 8000x - 4000 = 0 can be factored as:
(x - 100)(x + 40) ≤ 0
The critical points for this inequality are x = -40 and x = 100.
Let's test the intervals:
For x < -40, both factors are negative, so the inequality is not satisfied.
For -40 ≤ x ≤ 100, one factor is negative and the other is positive, so the inequality is satisfied.
For x > 100, both factors are positive, so the inequality is not satisfied.
Therefore, the range of rates at which customers can arrive before a second attendant is needed is -40 ≤ x ≤ 100. In interval form, this range is (-40, 100].
To determine the range of rates at which customers can arrive before a second attendant is needed, we need to find the values of x for which the average line length (F(x)) does not exceed 10 customers.
Given that F(x) = (x^2/400) - 20x, we can set F(x) = 10 and solve for x.
(x^2/400) - 20x = 10
Multiplying both sides of the equation by 400, we get:
x^2 - 4000x - 4000 = 0
This is a quadratic equation. We can solve it by factoring or using the quadratic formula.
Factoring the quadratic equation, we get:
(x - 100)(x + 40) = 0
This gives us two possible solutions: x = 100 and x = -40.
However, since the given range for x is 0 <= x < 20, we discard the solution x = 100 (which lies outside the given range).
So, the only valid solution is x = -40, but this is not a valid rate for customer arrival.
Therefore, there is no rate at which customers can arrive before a second attendant is needed within the given constraints.
In interval form, the answer would be an empty set: ∅.