We have that
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.
