Notes: The yearly profit of a company in thousands of dollars can be modeled by P(t) = t4 − 5t2 + 4, where t is the number of years since...

Monday, 13 March 2017

The yearly profit of a company in thousands of dollars can be modeled by P(t) = t4 − 5t2 + 4, where t is the number of years since...

We are given $\displaystyle{P}{\left({t}\right)}={{t}}^{{4}}-{5}{{t}}^{{2}}+{4}$ as a model for the profits earned, where t is the time in years since 2005. We are asked to find the years with zero profit.

This is equivalent to solving the equation $\displaystyle{{t}}^{{4}}-{5}{{t}}^{{2}}+{4}={0}$ .

Note that this is a quadratic form -- it is a quadratic in $\displaystyle{{t}}^{{2}}$ .

$\displaystyle{{t}}^{{4}}-{5}{{t}}^{{2}}+{4}={\left({{t}}^{{2}}-{4}\right)}{\left({{t}}^{{2}}-{1}\right)}$

Each of the binomials is a difference of two squares and can be factored as:

$\displaystyle={\left({t}+{2}\right)}{\left({t}-{2}\right)}{\left({t}+{1}\right)}{\left({t}-{1}\right)}\ldots$

This is equivalent to solving the equation $\displaystyle{{t}}^{{4}}-{5}{{t}}^{{2}}+{4}={0}$ .

Note that this is a quadratic form -- it is a quadratic in $\displaystyle{{t}}^{{2}}$ .

$\displaystyle{{t}}^{{4}}-{5}{{t}}^{{2}}+{4}={\left({{t}}^{{2}}-{4}\right)}{\left({{t}}^{{2}}-{1}\right)}$

Each of the binomials is a difference of two squares and can be factored as:

$\displaystyle={\left({t}+{2}\right)}{\left({t}-{2}\right)}{\left({t}+{1}\right)}{\left({t}-{1}\right)}$

So (t+2)(t-2)(t+1)(t-1)=0

$\displaystyle{t}=\pm{1},\pm{2}$

Since the domain is positive (the model is true for years after 2005), t=1 is 2006 and t=2 is 2007.

-----------------------------------------------------------------------------

The profit was zero in 2006 and 2007

-----------------------------------------------------------------------------

There were 4 solutions, but 2 of them (-1 and -2) were not in the domain and so are disregarded. (If the company came into existence in 2005, it makes no sense to describe its profits before 2005.)

The graph for t>0: