Respuesta :
Answer:
[tex] P'(t) = 3.2 t -11[/tex]
And we can set equal this derivate to 0 in order to find the critical point and we got:
[tex] 3.2 t-11= 0[/tex]
[tex] t = \frac{11}{3.2}= 3.4375[/tex]
And we can calculate the second derivate and we got:
[tex]P''(t) = 3.2 >0[/tex]
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:
[tex] P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094[/tex]
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:
[tex] P(t) = 1.6 t^2 -11t +44[/tex]
Where P represent the annual net income for the period 2007-2011 and [tex] 2 \leq t \leq 7[/tex]
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:
[tex] P'(t) = 3.2 t -11[/tex]
And we can set equal this derivate to 0 in order to find the critical point and we got:
[tex] 3.2 t-11= 0[/tex]
[tex] t = \frac{11}{3.2}= 3.4375[/tex]
And we can calculate the second derivate and we got:
[tex]P''(t) = 3.2 >0[/tex]
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:
[tex] P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094[/tex]
So then the minimum annual income occurs at t = 3.43 (between 2008 and 2009) and the value is 25.094
