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

Question

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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?

Mathematics

1 answer:

babunello [8.4K] · Answer 1 · 2026-03-22T13:13:59+03:00

Answer:

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

To identify the critical point, we set the derivative equal to zero, resulting in:

$3.2 t-11= 0$

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

Calculating the second derivative yields:

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

This indicates that t = 3.4375 marks the minimum value for the function. Substituting this back into the original function gives us:

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

Thus, the minimum annual income is found at t = 3.43 (between the years 2008 and 2009), with a value of 25.094

Step-by-step explanation:

In this scenario, we have the function:

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

Where P represents the annual net income from 2007-2011, and $2 \leq t \leq 7$

t is the number of years since early 2005

To discover the lowest income, we utilize the derivative given by:

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

Setting this derivative to zero allows us to find the critical point, leading to:

$3.2 t-11= 0$

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

Calculating the second derivative reveals:

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

Therefore, we conclude that t = 3.4375 is the minimum value, substituting into the original function results in:

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

Thus, the minimum annual income occurs at t = 3.43 (between 2008 and 2009) with the value being 25.094