Question

The number of bacteria, measured in thousands, in a culture is modelled by the equation $ b(t)=\dfrac{380{{e}^{2.31t}}}{175+{{e}^{2.31t}}} $ , where t is the number of days since the culture was formed. According to this model. The culture can support a maximum population of
A. 217
B. 205
C. 380
D. 760
E. $ \infty $

Answer 1

Hint: When a given function (here it is b) is at its maximum value, the first derivative of the function with respect to the variable on which the function is dependent (in this case it is t) is equal to zero and the second the derivative of the same function with respect to the variable is negative

Complete step-by-step answer:
It is given that the population of a certain bacteria as $ b(t)=\dfrac{380{{e}^{2.31t}}}{175+{{e}^{2.31t}}} $ ….. (i), where b is the population of the bacteria (measured in thousands) at a given time t.
Here, we can see that ‘b’ is a function of time t. When a given function (here it is b) is at its maximum value, the first derivative of the function with respect to the variable on which the function is dependent (in this case it is t) is equal to zero and the second the derivative of the same function with respect to the variable is negative.
Therefore, let us find the first derivative of b(t) with respect to t.
We can write (i) as $ \left( 175+{{e}^{2.31t}} \right)b=380{{e}^{2.31t}} $ .
Now, differentiate the above equation with respect to t.
 $ \Rightarrow \left( 175+{{e}^{2.31t}} \right)b'+\left( 2.31{{e}^{2.31t}} \right)b=380(2.31){{e}^{2.31t}} $ … (*).
 $ \Rightarrow b'=\dfrac{380(2.31){{e}^{2.31t}}-\left( 2.31{{e}^{2.31t}} \right)b}{\left( 175+{{e}^{2.31t}} \right)} $ …. (ii).
Now, put $ b'=0 $ .
 $ \Rightarrow 0=\dfrac{380(2.31){{e}^{2.31t}}-\left( 2.31{{e}^{2.31t}} \right)b}{\left( 175+{{e}^{2.31t}} \right)} $
 $ \Rightarrow 0=380(2.31){{e}^{2.31t}}-\left( 2.31{{e}^{2.31t}} \right)b $
 $ \Rightarrow b=380 $ .
We can write equation (ii) as
 $ \Rightarrow b'=(2.31)\left[ \dfrac{380{{e}^{2.31t}}}{\left( 175+{{e}^{2.31t}} \right)}-\dfrac{\left( {{e}^{2.31t}} \right)b}{\left( 175+{{e}^{2.31t}} \right)} \right] $
With the help of equation with can simplify the equation to:
 $ \Rightarrow b'=(2.31)\left[ b-\dfrac{{{b}^{2}}}{380} \right] $
Now, differentiate the equation with respect to.
 $ \Rightarrow b''=(2.31)\left[ b'-\dfrac{2bb'}{380} \right] $
At the point, $ b'=0 $ .
 $ \Rightarrow b''=(2.31)\left[ 0-\dfrac{2b(0)}{380} \right] $
 $ \Rightarrow b''=0 $ .
This means that the second derivative of the function b(t) is zero. And if we go on to find the further derivatives of the functions, they will be zero when $ b=380 $ . This means that from the point where $ b=380 $ , function b(t) remains constant and equal to 380.
Therefore, we can say that at time $ t=0 $ , the population of bacteria is less than 80, it increases to 380 after some days and remains constant from that time. Hence the maximum population of the bacteria is 380 thousand.
So, the correct answer is “Option C”.

Note: To find the first derivative of the function b(t), we can use the quotient rule of differentiation on equation (ii). However, using product rule is easier than the quotient rule that is why we wrote it is the form that shown in (*).