Question

The annual net income of a company for the period 2007–2011 could be approximated by P(t) = 1.6t2 − 11t + 44 billion dollars (2 ≤ t ≤ 6), where t is the time in years since the start of 2005.
According to the model, during what year in this period was the company's net income the lowest?

Answer 1

Answer:

$P'(t) = 3.2 t -11$

And we can set equal this derivate to 0 in order to find the critical point and we got:

$3.2 t-11= 0$

$t = \frac{11}{3.2}= 3.4375$

And we can calculate the second derivate and we got:

$P''(t) = 3.2 >0$

So then w can conclude that the value of t = 3.4375 represent the minimum value for the function and we can replace in the original function and we got:

$P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094$

So then the minimum annual income occurs at t = 3.43 (between 2008 and 2009) and the value is 25.094

Step-by-step explanation:

For this case we have the following function:

$P(t) = 1.6 t^2 -11t +44$

Where P represent the annual net income for the period 2007-2011 and $2 \leq t \leq 7$

And t represent the time in years since the start of 2005

In order to find the lowet income we need to use the derivate, given by:

$P'(t) = 3.2 t -11$

And we can set equal this derivate to 0 in order to find the critical point and we got:

$3.2 t-11= 0$

$t = \frac{11}{3.2}= 3.4375$

And we can calculate the second derivate and we got:

$P''(t) = 3.2 >0$

So then w can conclude that the value of t = 3.4375 represent the minimum value for the function and we can replace in the original function and we got:

$P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094$

So then the minimum annual income occurs at t = 3.43 (between 2008 and 2009) and the value is 25.094