An online retail company determined that their cost for each day, x, can be determined by the function C(x) = 2x2 – 20x + 80 and their revenue by the function R(x) = –x2 + 5x + 58. How many days will it take for the company to earn a profit?



Answer :

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.

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