A cell phone company’s profit is modeled by the function p(x) = 10x − x2 − 1, where p is expressed in thousands of dollars and x is the number of phones sold in hundreds. How many phones must the company sell to make a profit of more than $15,000?
P(x) = 10x - x^2 - 1 = 15.
-x^2 + 10x - 1 - 15 = 0.
-x^2 + 10x - 16 = 0.
Solve using Quadratic Formula and get:
X = 2 = 200 Phones.
X = 8 = 800 Phnes.
The profit, P(x), of a software company, in thousands of dollars, is given by the equation P(x) = "-10x+" 110x + 10000, where x is the amount spent on advertising, in thousands of dollars. Determine the least amount spent on advertising that will result in a profit of at least $8 000 000.
To find the number of phones the company must sell to make a profit of more than $15,000, we need to set up an inequality using the given profit function.
The profit function is given as p(x) = 10x - x^2 - 1, where p is in thousands of dollars and x is the number of phones sold in hundreds.
We want to find the value of x when the profit is greater than $15,000, which is expressed as 15,000 in thousands of dollars.
So we can write the inequality as follows:
10x - x^2 - 1 > 15
Rearranging the inequality, we have:
x^2 - 10x + 15 < 0
Factoring the quadratic equation, we get:
(x - 5)(x - 3) < 0
To solve this inequality, we need to find the critical points, which are the values that make the expression equal to zero. In this case, the critical points are x = 5 and x = 3.
Now we can set up a sign chart:
| - | + | - |
_______________________
| x | x | x |
_______________________
- 3 | 5 | ∞
In the intervals between the critical points, we test a value to determine the sign of the expression:
For x < 3, we can test x = 2:
(2 - 5)(2 - 3) = (-3)(-1) = 3 > 0
For 3 < x < 5, we can test x = 4:
(4 - 5)(4 - 3) = (-1)(1) = -1 < 0
For x > 5, we can test x = 6:
(6 - 5)(6 - 3) = (1)(3) = 3 > 0
Since we are looking for values of x where the expression is less than zero, we are interested in the intervals where the expression is negative.
From the sign chart, we can see that the intervals where the expression is negative are:
x < 3 and 5 < x < ∞
Therefore, the number of phones the company must sell to make a profit of more than $15,000 is within the range x < 3 or x > 5.
In terms of hundreds of phones, the company must sell fewer than 300 phones or more than 500 phones to achieve a profit of more than $15,000.
To find out how many phones the company must sell to make a profit of more than $15,000, we need to solve the inequality 10x - x^2 - 1 > 15.
The equation for the profit is p(x) = 10x - x^2 - 1.
Step 1: Set up the inequality:
10x - x^2 - 1 > 15
Step 2: Move all terms to one side of the inequality:
10x - x^2 - 1 - 15 > 0
- x^2 + 10x - 16 > 0
Step 3: Simplify the equation:
-x^2 + 10x - 16 > 0
Now, we have a quadratic inequality in the form ax^2 + bx + c > 0.
Step 4: Solve the quadratic equation:
Since the quadratic equation is not easily factored, we can use the quadratic formula.
The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a).
For our equation, a = -1, b = 10, and c = -16.
x = (-10 ± √(10^2 - 4(-1)(-16))) / (2(-1))
x = (-10 ± √(100 - 64)) / (-2)
x = (-10 ± √36) / (-2)
x = (-10 ± 6) / (-2)
Step 5: Simplify the solutions:
We have two solutions: (-10 + 6) / (-2) and (-10 - 6) / (-2).
The simplified solutions are x = -2 and x = 8.
Step 6: Determine the correct solution:
Since we are concerned with the number of phones sold, the value of x cannot be negative. Therefore, we discard the solution x = -2.
So, the company must sell more than 800 phones (8 x 100) to make a profit of over $15,000.