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Sagot :
### Part B: Solving the Questions Step-by-Step
#### Question 3: Use technology or hand calculations to determine the equation for the quadratic function modeled by the data in the table.
Let's start with the given data points:
[tex]\[ \begin{array}{|c|c|} \hline \text{Number of Price Increases } (x) & \text{Profit (millions) } (y) \\ \hline 25 & 204 \\ 60 & 285 \\ 105 & 368 \\ 132 & 384 \\ 155 & 376 \\ 172 & 356 \\ 200 & 318 \\ 254 & 185 \\ 290 & 70 \\ \hline \end{array} \][/tex]
We find a quadratic function of the form [tex]\( y = ax^2 + bx + c \)[/tex]. After fitting the quadratic model to the data points using regression analysis, we get the equation:
[tex]\[ y = 117.91363072818619 + 3.742764671615785x - 0.013564595957476968x^2 \][/tex]
or more neatly:
[tex]\[ y = 117.91 + 3.74x - 0.0136x^2 \][/tex]
This is the quadratic equation fitting the given data.
#### Question 4: Using the equation from question 3, determine the maximum profit.
For a quadratic function [tex]\(y = ax^2 + bx + c\)[/tex], the vertex of the parabola, which corresponds to the maximum (or minimum) value, occurs at [tex]\(x = -\frac{b}{2a}\)[/tex].
In our equation:
[tex]\[ y = 117.91363072818619 + 3.742764671615785x - 0.013564595957476968x^2 \][/tex]
Here, [tex]\(a = -0.013564595957476968\)[/tex], [tex]\(b = 3.742764671615785\)[/tex].
To find the [tex]\(x\)[/tex]-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} = -\frac{3.742764671615785}{2 \times -0.013564595957476968} \approx 137.96078716051724 \][/tex]
Substitute [tex]\( x = 137.96078716051724 \)[/tex] back into the quadratic equation to find the maximum profit [tex]\( y \)[/tex]:
[tex]\[ y = 117.91363072818619 + 3.742764671615785 \times 137.96078716051724 - 0.013564595957476968 \times (137.96078716051724)^2 \approx 376.09101085453045 \][/tex]
Thus, the maximum profit is approximately [tex]\(\$376.09 \)[/tex] million.
#### Question 5: Using the equation from question 3, determine how many price increases will cause the smoothie chain to have zero profit.
To find the number of price increases that result in zero profit, we need to solve the quadratic equation [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 117.91363072818619 + 3.742764671615785x - 0.013564595957476968x^2 \][/tex]
This simplifies to:
[tex]\[ -0.013564595957476968x^2 + 3.742764671615785x + 117.91363072818619 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = -0.013564595957476968, \quad b = 3.742764671615785, \quad c = 117.91363072818619 \][/tex]
[tex]\[ x = \frac{-3.742764671615785 \pm \sqrt{(3.742764671615785)^2 - 4 \times (-0.013564595957476968) \times 117.91363072818619}}{2 \times -0.013564595957476968} \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = (3.742764671615785)^2 - 4 \times (-0.013564595957476968) \times 117.91363072818619 \approx 137.29120756665236 \][/tex]
Thus:
[tex]\[ x = \frac{-3.742764671615785 \pm \sqrt{137.29120756665236}}{2 \times -0.013564595957476968} \][/tex]
Solving this, we get the two solutions:
[tex]\[ x \approx -28.5502575156461 \quad \text{and} \quad x \approx 304.471831836680 \][/tex]
Therefore, the number of price increases that will result in zero profit are approximately:
[tex]\[ x \approx -28.55 \quad \text{and} \quad x \approx 304.47 \][/tex]
Since a negative number of price increases doesn't make sense in this context, we take the positive solution:
[tex]\[ 304.47 \][/tex]
So, approximately 304.47 price increases will result in zero profit.
#### Question 3: Use technology or hand calculations to determine the equation for the quadratic function modeled by the data in the table.
Let's start with the given data points:
[tex]\[ \begin{array}{|c|c|} \hline \text{Number of Price Increases } (x) & \text{Profit (millions) } (y) \\ \hline 25 & 204 \\ 60 & 285 \\ 105 & 368 \\ 132 & 384 \\ 155 & 376 \\ 172 & 356 \\ 200 & 318 \\ 254 & 185 \\ 290 & 70 \\ \hline \end{array} \][/tex]
We find a quadratic function of the form [tex]\( y = ax^2 + bx + c \)[/tex]. After fitting the quadratic model to the data points using regression analysis, we get the equation:
[tex]\[ y = 117.91363072818619 + 3.742764671615785x - 0.013564595957476968x^2 \][/tex]
or more neatly:
[tex]\[ y = 117.91 + 3.74x - 0.0136x^2 \][/tex]
This is the quadratic equation fitting the given data.
#### Question 4: Using the equation from question 3, determine the maximum profit.
For a quadratic function [tex]\(y = ax^2 + bx + c\)[/tex], the vertex of the parabola, which corresponds to the maximum (or minimum) value, occurs at [tex]\(x = -\frac{b}{2a}\)[/tex].
In our equation:
[tex]\[ y = 117.91363072818619 + 3.742764671615785x - 0.013564595957476968x^2 \][/tex]
Here, [tex]\(a = -0.013564595957476968\)[/tex], [tex]\(b = 3.742764671615785\)[/tex].
To find the [tex]\(x\)[/tex]-coordinate of the vertex:
[tex]\[ x = -\frac{b}{2a} = -\frac{3.742764671615785}{2 \times -0.013564595957476968} \approx 137.96078716051724 \][/tex]
Substitute [tex]\( x = 137.96078716051724 \)[/tex] back into the quadratic equation to find the maximum profit [tex]\( y \)[/tex]:
[tex]\[ y = 117.91363072818619 + 3.742764671615785 \times 137.96078716051724 - 0.013564595957476968 \times (137.96078716051724)^2 \approx 376.09101085453045 \][/tex]
Thus, the maximum profit is approximately [tex]\(\$376.09 \)[/tex] million.
#### Question 5: Using the equation from question 3, determine how many price increases will cause the smoothie chain to have zero profit.
To find the number of price increases that result in zero profit, we need to solve the quadratic equation [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 117.91363072818619 + 3.742764671615785x - 0.013564595957476968x^2 \][/tex]
This simplifies to:
[tex]\[ -0.013564595957476968x^2 + 3.742764671615785x + 117.91363072818619 = 0 \][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = -0.013564595957476968, \quad b = 3.742764671615785, \quad c = 117.91363072818619 \][/tex]
[tex]\[ x = \frac{-3.742764671615785 \pm \sqrt{(3.742764671615785)^2 - 4 \times (-0.013564595957476968) \times 117.91363072818619}}{2 \times -0.013564595957476968} \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = (3.742764671615785)^2 - 4 \times (-0.013564595957476968) \times 117.91363072818619 \approx 137.29120756665236 \][/tex]
Thus:
[tex]\[ x = \frac{-3.742764671615785 \pm \sqrt{137.29120756665236}}{2 \times -0.013564595957476968} \][/tex]
Solving this, we get the two solutions:
[tex]\[ x \approx -28.5502575156461 \quad \text{and} \quad x \approx 304.471831836680 \][/tex]
Therefore, the number of price increases that will result in zero profit are approximately:
[tex]\[ x \approx -28.55 \quad \text{and} \quad x \approx 304.47 \][/tex]
Since a negative number of price increases doesn't make sense in this context, we take the positive solution:
[tex]\[ 304.47 \][/tex]
So, approximately 304.47 price increases will result in zero profit.
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