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Answer:
2.119 days
Step-by-step explanation:
A profit is earned when the amount earned by a business exceeds the amount it has spent. Symbolically, this is:
[tex]\displaystyle \int_0^T R(x)\ dx > \int_0^T C(x)\ dx[/tex]
at time [tex]T[/tex] after the business starts where total amount earned is an integral of revenue (the current amount received), and amount spent is an integral of cost (the current price of operation).
To determine the initial point when the company makes a profit, we need to use the second part of the fundamental theorem of calculus and make a sign chart for the integrated functions:
[tex]\displaystyle \int_{a}^b f(x) \ dx = \left[\dfrac{}{}F(x)\dfrac{}{}\right]_a^b \ \ \ \ \ \ \text{where} \ \ f(x) = F'(x) \\ \\ \text{ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ = F(b)-F(a)[/tex]
↓↓↓
[tex]\displaystyle \int_0^T C(x) \ dx = \int_0^T (2x^2 - 20x + 80)\ dx[/tex]
[tex]=\left[\ \dfrac{2}{3}x^3 - 10x^2 + 80x\ \right]_0^T[/tex]
[tex]=\dfrac{2}{3}T^3 - 10T^2 + 80T[/tex]
__
[tex]\displaystyle \int_0^T R(x) \ dx = \int_0^T (-x^2 + 5x + 58)\ dx[/tex]
[tex]= \left[\ -\dfrac{1}{3}x^3 + \dfrac{5}{2}x^2 + 58x\ \right]_0^T[/tex]
[tex]= -\dfrac{1}{3}T^3 + \dfrac{5}{2}T^2 + 58T[/tex]
Now, we look at the inequality:
[tex]-\dfrac{1}{3}T^3 + \dfrac{5}{2}T^2 + 58T\ > \ \dfrac{2}{3}T^3 - 10T^2 + 80T[/tex]
↓ getting all the terms on one side
[tex]-T^3 + 12.5T^2 - 22T > 0[/tex]
We can create a sign chart with the roots of this inequality using a graphical calculator. This gives us approximately:
2.119 days.
