The profit function for a computer company is given by P(x)=−x^2+36x−26 where x is the number of units produced (in thousands) and the profit is in thousand of dollars.
a) Determine how many (thousands of) units must be produced to yield maximum profit. Determine the maximum profit.
(thousands of) units =
maximum profit = thousand dollars
b) Determine how many units should be produced for a profit of at least 40 thousand.
more than (thousands of) units
Thanks so much!
less than (thousands of) units
take the derivative, set it equal to zero and solve for x
Plug that x back into P(x) to get your maximum profit.
b) −x^2+36x−26 ≥ 40
x^2 - 36x + 66 ≤ 0
let's use the = sign
x^2 - 36x + 66 = 0
use the quadratic formula, you will get two positive x's
lower ≤ x ≤ larger
verify your answer by looking at this graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+%E2%88%92x%5E2%2B36x%E2%88%9226
a) To determine the number of thousands of units that must be produced to yield maximum profit, we need to find the x-value of the vertex of the profit function. The x-value of the vertex can be found using the formula: x = -b / (2a), where the profit function is in the form P(x) = ax^2 + bx + c.
In this case, the profit function is P(x) = -x^2 + 36x - 26. Comparing this to the general form, we can see that a = -1, b = 36, and c = -26.
Using the formula, we can calculate x = -36 / (2 * -1) = 18.
So, the number of thousands of units that must be produced to yield maximum profit is 18.
To find the maximum profit, we substitute the value of x into the profit function:
P(18) = -(18)^2 + 36(18) - 26 = -324 + 648 - 26 = 298 thousand dollars.
Therefore, the maximum profit is 298 thousand dollars.
b) To determine how many units should be produced for a profit of at least 40 thousand dollars, we need to solve the profit function for P(x) = 40.
- x^2 + 36x - 26 = 40
Rearranging the equation, we get:
x^2 - 36x + 66 = 0
Unfortunately, this equation does not have real solutions, meaning there is no integer value of x that will give us a profit of at least 40 thousand dollars.
Therefore, there is no specific number of units that should be produced to achieve a profit of at least 40 thousand dollars.
a) To find the number of units that must be produced to yield maximum profit, we need to find the x-value of the vertex of the quadratic function. The vertex can be found using the formula: x = -b / (2a), where a = -1 and b = 36.
x = -36 / (2 * -1)
x = -36 / -2
x = 18
Therefore, the number of thousands of units that must be produced to yield maximum profit is 18 thousand units.
To find the maximum profit, we substitute the x-value into the profit function:
P(18) = -(18^2) + 36(18) - 26
P(18) = -324 + 648 - 26
P(18) = 298
Therefore, the maximum profit is 298 thousand dollars.
b) To determine the number of units that should be produced for a profit of at least 40 thousand, we set the profit function greater than or equal to 40:
P(x) ≥ 40
Now we solve this inequality by rearranging it to find the range of x:
-x^2 + 36x - 26 ≥ 40
-x^2 + 36x - 26 - 40 ≥ 0
-x^2 + 36x - 66 ≥ 0
We can solve this quadratic inequality using factoring or the quadratic formula, but it turns out that in this case, the quadratic does not have real roots. This means that the quadratic function is always negative and never greater than or equal to zero. Therefore, there are no units that can be produced to achieve a profit of at least 40 thousand.
Therefore, the answer is: There are no units that should be produced for a profit of at least 40 thousand.
To find the number of units (in thousands) that must be produced to yield the maximum profit, we need to find the x-value at which the profit function P(x) is maximized. This can be achieved by finding the vertex of the quadratic function.
The profit function is given by P(x) = -x^2 + 36x - 26. To find the vertex, we can use the formula x = -b / (2a), where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c.
In this case, a = -1, b = 36, and c = -26. Substituting these values into the formula, we get:
x = -36 / (2 * -1)
x = -36 / -2
x = 18
So, the number of units (in thousands) that must be produced to yield maximum profit is 18.
To find the maximum profit, we can substitute the value of x into the profit function:
P(18) = -(18)^2 + 36(18) - 26
P(18) = -324 + 648 - 26
P(18) = 298
Therefore, the maximum profit is 298 thousand dollars.
To determine how many units should be produced for a profit of at least 40 thousand, we need to find the x-value(s) for which the profit function is greater than or equal to 40.
Setting P(x) ≥ 40, we have:
-x^2 + 36x - 26 ≥ 40
Simplifying, we get:
-x^2 + 36x - 26 - 40 ≥ 0
-x^2 + 36x - 66 ≥ 0
To solve this quadratic inequality, we can either use factoring, graphing, or the quadratic formula. Since factoring is not feasible in this case, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -1, b = 36, and c = -66. Substituting these values into the quadratic formula, we get:
x = (-36 ± √(36^2 - 4(-1)(-66))) / (2 * -1)
x = (-36 ± √(1296 - 264)) / -2
x = (-36 ± √1032) / -2
Simplifying further, we have:
x ≈ (-36 + √1032) / -2
x ≈ (-36 + 32.12) / -2
x ≈ (-3.88) / -2
x ≈ 1.94
Therefore, the number of units (in thousands) that should be produced for a profit of at least 40 thousand is more than 1.94 thousand units.
Conversely, if we want to determine the number of units (in thousands) that should be produced for a profit of less than 40 thousand, we need to find the x-value(s) for which the profit function is less than 40. We can follow a similar process as above and find the appropriate x-values.