An epidemiologist determines that a particular epidemic spreads in such a way that t weeks after the outbreak, N hundred new cases will be reported where N(t)= 5t/(12+t*2). At what time is the epidemic at its worst? What is the maximum number of reported new cases?

The epidemic is a maximum when

dN/dt = 0

dN/dt = [-5t*(2t) + 5(12 + t^2)]/(12 + t^2) = 0
That will be zero when the numerator is zero, so
-10t^2 + 60 + 5 t^2 = 0
t^2 = 12
t = 3.46 weeks

N(3.46) = 17.3/23.97 = 0.748 hundred new cases (75 cases)

To determine the time when the epidemic is at its worst and the maximum number of reported new cases, we need to find the maximum value of the function N(t) = 5t/(12 + t^2).

Step 1: Let's first find the derivative of N(t) with respect to t:

N'(t) = dN/dt = (d/dt)(5t/(12 + t^2))

To find the derivative, we can use the quotient rule:

N'(t) = (5(12 + t^2) - 5t(2t))/(12 + t^2)^2

Simplifying this expression:

N'(t) = (60 + 5t^2 - 10t^2)/(12 + t^2)^2

N'(t) = (60 - 5t^2)/(12 + t^2)^2

Step 2: To find the critical points, we need to solve the equation N'(t) = 0:

(60 - 5t^2)/(12 + t^2)^2 = 0

Since the numerator can't equal zero (60 - 5t^2 ≠ 0), there are no critical points.

Step 3: We need to check the endpoints of the interval where t is defined. Since the function is undefined at t = ±√12, we can exclude those values.

Step 4: We can determine the behavior of the function as t approaches infinity by examining the limits. Let's find the limit as t approaches infinity:

lim (t→∞) (5t/(12 + t^2))

To calculate the limit, we divide every term in the numerator and denominator by t^2:

lim (t→∞) (5/t)/(12/t^2 + 1)

As t approaches infinity, the fraction (5/t) approaches 0, and the fraction (12/t^2) approaches 0:

lim (t→∞) 0/(0 + 1) = 0/1 = 0

Step 5: Since we have no critical points and the limit as t approaches infinity is 0, there are no local maximum or minimum values for N(t).

Therefore, the epidemic does not have a specific time when it is at its worst, and there is no maximum number of reported new cases.

To find the time when the epidemic is at its worst and the maximum number of reported new cases, we need to find the maximum value of the function N(t) = 5t/(12 + t^2).

To determine the maximum value, we can take the derivative of N(t) with respect to t and set it equal to zero. The derivative will give us the rate of change of N(t) with respect to time (t).

Let's find the derivative of N(t):

N'(t) = [d/dt(5t)] / (12 + t^2) + [d/dt(12 + t^2)] * (5t)

Simplifying further:

N'(t) = (5) / (12 + t^2) + 0 - (10t^2) / (12 + t^2)^2

Now, set N'(t) equal to zero and solve for t:

(5) / (12 + t^2) - (10t^2) / (12 + t^2)^2 = 0

Multiplying through by (12 + t^2)^2:

5(12 + t^2) - 10t^2 = 0

60 + 5t^2 - 10t^2 = 0

-5t^2 + 60 = 0

Dividing through by -5:

t^2 - 12 = 0

t^2 = 12

Taking the square root of both sides:

t = ±√12

Since we're interested in the time when the epidemic is at its worst, we can disregard the negative solution. So, t = √12.

Now, substitute the value of t = √12 into the function N(t) to find the maximum number of reported new cases:

N(√12) = 5(√12) / (12 + (√12)^2)

Simplifying further:

N(√12) = 5√12 / (12 + 12)

N(√12) = (5√12) / 24

Therefore, the time when the epidemic is at its worst is t = √12, and the maximum number of reported new cases is (5√12) / 24.