Answer: 3rd year of operation
Explanation:
Note that p(t) is a quadratic function and so its graph is a parabola. Since the coefficient of t² in p(t) is negative, the maximum point in p(t) exist at its vertex. Moreover, the maximum value of p(t) is the y-coordinate of its vertex.
Note that p(t) can be expressed as:
[tex]p(t) = a(t-h)^2 + k [/tex] (1)
Where (h, k) are the coordinates of the vertex of p(t).
To manipulate p(t) in the form expressed in equation (1), we factor out the coefficient of t² in p(t) so that
[tex]p(t) = -3t^2+18t-4
\\ \boxed{p(t) = -3 \left( t^2 - 6t + \frac{4}{3} \right)}[/tex]
Then, we let
[tex]q(t) = t^2 - 6t + \frac{4}{3}[/tex]
So that
[tex]p(t) = -3q(t)[/tex] (2)
We need q(t) to be expressed as the sum of a perfect square trinomial and a constant. To form the perfect square trinomial in q(t), we can find a constant k such that [tex]t^2 - 6t + k [/tex] is a perfect square.
To find the value of k, we divide the coefficient of t by 2 and get the square of the result. Since the coefficient of t in q(t) is -6,
[tex]k = \left( \frac{-6}{2} \right)^2 = 9[/tex]
To avoid changing the value of q(t), if we add the constant k = 9, we need to subtract it by the same number. Since k = 9,
[tex]q(t) = t^2 - 6t + \frac{4}{3} + k - k
\\ = t^2 - 6t + \frac{4}{3} + 9 - 9
\\ = (t^2 - 6t + 9) + \frac{4}{3} - 9
\\ \boxed{q(t) = (t - 3)^2 - \frac{23}{3}} [/tex]
From equation (2),
[tex]p(t) = -3q(t)
\\ p(t) = -3\left( (t - 3)^2 - \frac{23}{3} \right)
\\ \boxed{p(t) = -3 (t - 3)^2 + 23}[/tex]
Hence the vertex of p(t) is (3, 23) and maximum value is attained at t = 3. Therefore, the Mr. Cash's business has maximum profit at the 3rd year of operation.
