Answer :
Let's determine the number of dolls sold for which the company's income equals its costs. We need to find where the two equations intersect.
The equations are:
Income: [tex]\( y = -0.4x^2 + 3x + 45 \)[/tex]
Costs: [tex]\( y = 1.5x + 20 \)[/tex]
To find the break-even point(s), we set the income equal to the costs:
[tex]\[ -0.4x^2 + 3x + 45 = 1.5x + 20 \][/tex]
First, rearrange this equation to form a standard quadratic equation:
[tex]\[ -0.4x^2 + 3x - 1.5x + 45 - 20 = 0 \][/tex]
[tex]\[ -0.4x^2 + 1.5x + 25 = 0 \][/tex]
In the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\( a = -0.4 \)[/tex]
[tex]\( b = 1.5 \)[/tex]
[tex]\( c = 25 \)[/tex]
Next, calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{Discriminant} = (1.5)^2 - 4(-0.4)(25) \][/tex]
A quadratic equation can have:
- Two solutions if the discriminant is greater than 0
- One solution if the discriminant equals 0
- No real solutions if the discriminant is less than 0
In this case:
[tex]\[ \text{Discriminant} = 2.25 + 40 = 42.25 \][/tex]
Since the discriminant [tex]\( 42.25 \)[/tex] is greater than 0, there are 2 potential solutions for [tex]\( x \)[/tex].
However, we need to determine which of these solutions are viable. In real-world scenarios like this, [tex]\( x \)[/tex] must be non-negative because it represents the number of dolls sold.
By calculating the roots of the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{-1.5 \pm \sqrt{42.25}}{-0.8} \][/tex]
[tex]\[ x_1 = \frac{5.75}{-0.8} \implies x_1 < 0 \][/tex] (Not viable)
[tex]\[ x_2 = \frac{-8.75}{-0.8} \implies x_2 > 0 \][/tex] (Viable)
Here, [tex]\( x_1 \)[/tex] is a negative number which is not meaningful in this context, but [tex]\( x_2 \)[/tex] is positive. Therefore, we have 1 viable solution.
Hence,
- There are 2 total possible solutions.
- There is 1 viable solution.
So the correct answers are:
- How many total possible solutions of the form ( [tex]\( x, y \)[/tex] ) are there for this situation? [tex]\(\boxed{2}\)[/tex]
- Of any possible solutions of the form ( [tex]\( x, y \)[/tex] ), how many are viable for this situation? [tex]\(\boxed{1}\)[/tex]
The equations are:
Income: [tex]\( y = -0.4x^2 + 3x + 45 \)[/tex]
Costs: [tex]\( y = 1.5x + 20 \)[/tex]
To find the break-even point(s), we set the income equal to the costs:
[tex]\[ -0.4x^2 + 3x + 45 = 1.5x + 20 \][/tex]
First, rearrange this equation to form a standard quadratic equation:
[tex]\[ -0.4x^2 + 3x - 1.5x + 45 - 20 = 0 \][/tex]
[tex]\[ -0.4x^2 + 1.5x + 25 = 0 \][/tex]
In the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], we have:
[tex]\( a = -0.4 \)[/tex]
[tex]\( b = 1.5 \)[/tex]
[tex]\( c = 25 \)[/tex]
Next, calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{Discriminant} = (1.5)^2 - 4(-0.4)(25) \][/tex]
A quadratic equation can have:
- Two solutions if the discriminant is greater than 0
- One solution if the discriminant equals 0
- No real solutions if the discriminant is less than 0
In this case:
[tex]\[ \text{Discriminant} = 2.25 + 40 = 42.25 \][/tex]
Since the discriminant [tex]\( 42.25 \)[/tex] is greater than 0, there are 2 potential solutions for [tex]\( x \)[/tex].
However, we need to determine which of these solutions are viable. In real-world scenarios like this, [tex]\( x \)[/tex] must be non-negative because it represents the number of dolls sold.
By calculating the roots of the quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ x = \frac{-1.5 \pm \sqrt{42.25}}{-0.8} \][/tex]
[tex]\[ x_1 = \frac{5.75}{-0.8} \implies x_1 < 0 \][/tex] (Not viable)
[tex]\[ x_2 = \frac{-8.75}{-0.8} \implies x_2 > 0 \][/tex] (Viable)
Here, [tex]\( x_1 \)[/tex] is a negative number which is not meaningful in this context, but [tex]\( x_2 \)[/tex] is positive. Therefore, we have 1 viable solution.
Hence,
- There are 2 total possible solutions.
- There is 1 viable solution.
So the correct answers are:
- How many total possible solutions of the form ( [tex]\( x, y \)[/tex] ) are there for this situation? [tex]\(\boxed{2}\)[/tex]
- Of any possible solutions of the form ( [tex]\( x, y \)[/tex] ), how many are viable for this situation? [tex]\(\boxed{1}\)[/tex]
