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Sagot :
Sure! Let's solve this step-by-step in a detailed fashion:
### Given:
We have a profit function:
[tex]\[ P = 4x^2 - 20x - 9 \][/tex]
Where:
- [tex]\( P \)[/tex] is the profit in millions of dollars.
- [tex]\( x \)[/tex] is the number of games produced in hundred thousands.
We need to find:
a) The break-even points (values of [tex]\( x \)[/tex] where [tex]\( P = 0 \)[/tex]).
b) The maximum profit.
c) The number of games that result in the maximum profit.
### a) Break-even points:
The break-even points occur where the profit [tex]\( P \)[/tex] is zero. Therefore, we need to solve the quadratic equation:
[tex]\[ 4x^2 - 20x - 9 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)[/tex], where [tex]\( a = 4 \)[/tex], [tex]\( b = -20 \)[/tex], and [tex]\( c = -9 \)[/tex]:
[tex]\[ x = \frac{{20 \pm \sqrt{{(-20)^2 - 4(4)(-9)}}}}{2(4)} \][/tex]
[tex]\[ x = \frac{{20 \pm \sqrt{{400 + 144}}}}{8} \][/tex]
[tex]\[ x = \frac{{20 \pm \sqrt{544}}}{8} \][/tex]
[tex]\[ x = \frac{{20 \pm 23.32}}{8} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{{20 + 23.32}}{8} \approx 5.4155 \][/tex]
[tex]\[ x_2 = \frac{{20 - 23.32}}{8} \approx -0.4155 \][/tex]
So, the break-even points are [tex]\( x \approx -0.4155 \)[/tex] and [tex]\( x \approx 5.4155 \)[/tex], where [tex]\( x \)[/tex] is in hundred thousands of games.
### b) Maximum profit:
To find the maximum profit, we need to find the vertex of the parabola defined by the quadratic function [tex]\( P = 4x^2 - 20x - 9 \)[/tex].
The x-coordinate of the vertex (where the maximum or minimum occurs) for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = \frac{-b}{2a} \][/tex]
In our case:
[tex]\[ a = 4 \][/tex]
[tex]\[ b = -20 \][/tex]
[tex]\[ x = \frac{20}{8} = 2.5 \][/tex]
So, the number of games (in hundred thousands) that results in maximum profit is [tex]\( x = 2.5 \)[/tex].
### c) Maximum profit calculation:
Substitute [tex]\( x = 2.5 \)[/tex] back into the profit function to find the maximum profit:
[tex]\[ P = 4(2.5)^2 - 20(2.5) - 9 \][/tex]
[tex]\[ P = 4(6.25) - 20(2.5) - 9 \][/tex]
[tex]\[ P = 25 - 50 - 9 \][/tex]
[tex]\[ P = -34 \][/tex]
Thus, the maximum profit is -34 million dollars.
### Summary:
a) The break-even points are at [tex]\( x \approx -0.4155 \)[/tex] and [tex]\( x \approx 5.4155 \)[/tex] (hundred thousands of games).
b) The maximum profit is -34 million dollars.
c) The number of games produced for the maximum profit is 2.5 hundred thousand games (or 250,000 games).
These results conclude the analysis of the profit function.
### Given:
We have a profit function:
[tex]\[ P = 4x^2 - 20x - 9 \][/tex]
Where:
- [tex]\( P \)[/tex] is the profit in millions of dollars.
- [tex]\( x \)[/tex] is the number of games produced in hundred thousands.
We need to find:
a) The break-even points (values of [tex]\( x \)[/tex] where [tex]\( P = 0 \)[/tex]).
b) The maximum profit.
c) The number of games that result in the maximum profit.
### a) Break-even points:
The break-even points occur where the profit [tex]\( P \)[/tex] is zero. Therefore, we need to solve the quadratic equation:
[tex]\[ 4x^2 - 20x - 9 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)[/tex], where [tex]\( a = 4 \)[/tex], [tex]\( b = -20 \)[/tex], and [tex]\( c = -9 \)[/tex]:
[tex]\[ x = \frac{{20 \pm \sqrt{{(-20)^2 - 4(4)(-9)}}}}{2(4)} \][/tex]
[tex]\[ x = \frac{{20 \pm \sqrt{{400 + 144}}}}{8} \][/tex]
[tex]\[ x = \frac{{20 \pm \sqrt{544}}}{8} \][/tex]
[tex]\[ x = \frac{{20 \pm 23.32}}{8} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{{20 + 23.32}}{8} \approx 5.4155 \][/tex]
[tex]\[ x_2 = \frac{{20 - 23.32}}{8} \approx -0.4155 \][/tex]
So, the break-even points are [tex]\( x \approx -0.4155 \)[/tex] and [tex]\( x \approx 5.4155 \)[/tex], where [tex]\( x \)[/tex] is in hundred thousands of games.
### b) Maximum profit:
To find the maximum profit, we need to find the vertex of the parabola defined by the quadratic function [tex]\( P = 4x^2 - 20x - 9 \)[/tex].
The x-coordinate of the vertex (where the maximum or minimum occurs) for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by:
[tex]\[ x = \frac{-b}{2a} \][/tex]
In our case:
[tex]\[ a = 4 \][/tex]
[tex]\[ b = -20 \][/tex]
[tex]\[ x = \frac{20}{8} = 2.5 \][/tex]
So, the number of games (in hundred thousands) that results in maximum profit is [tex]\( x = 2.5 \)[/tex].
### c) Maximum profit calculation:
Substitute [tex]\( x = 2.5 \)[/tex] back into the profit function to find the maximum profit:
[tex]\[ P = 4(2.5)^2 - 20(2.5) - 9 \][/tex]
[tex]\[ P = 4(6.25) - 20(2.5) - 9 \][/tex]
[tex]\[ P = 25 - 50 - 9 \][/tex]
[tex]\[ P = -34 \][/tex]
Thus, the maximum profit is -34 million dollars.
### Summary:
a) The break-even points are at [tex]\( x \approx -0.4155 \)[/tex] and [tex]\( x \approx 5.4155 \)[/tex] (hundred thousands of games).
b) The maximum profit is -34 million dollars.
c) The number of games produced for the maximum profit is 2.5 hundred thousand games (or 250,000 games).
These results conclude the analysis of the profit function.
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