To determine the number of dolls the company needs to sell to break-even, we need to set the company’s income equal to its costs and solve for [tex]\(x\)[/tex]. Here are the given equations:
Income:
[tex]\[ y = -0.4x^2 + 3x + 45 \][/tex]
Costs:
[tex]\[ y = 1.5x + 20 \][/tex]
To find the break-even points, we set the two equations equal to each other:
[tex]\[ -0.4x^2 + 3x + 45 = 1.5x + 20 \][/tex]
We then solve this equation for [tex]\(x\)[/tex]. This will give us the values of [tex]\(x\)[/tex] (the number of dolls sold in hundreds) at which income equals costs (break-even points).
Solving the equation:
[tex]\[ -0.4x^2 + 3x + 45 = 1.5x + 20 \][/tex]
Rearrange all terms to one side of the equation:
[tex]\[ -0.4x^2 + 3x - 1.5x + 45 - 20 = 0 \][/tex]
Simplify further:
[tex]\[ -0.4x^2 + 1.5x + 25 = 0 \][/tex]
Using the quadratic formula, [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our equation, [tex]\( a = -0.4 \)[/tex], [tex]\( b = 1.5 \)[/tex], and [tex]\( c = 25 \)[/tex].
Solving the equation gives us the roots or solutions for [tex]\(x\)[/tex]:
[tex]\[ x = -6.25 \quad \text{and} \quad x = 10 \][/tex]
These are the total possible solutions for [tex]\( x \)[/tex]. The break-even points occur at these values of [tex]\( x \)[/tex]. Therefore, there are 2 total possible solutions.
Now, we need to determine how many of these solutions are viable. Viability in this context means the number of dolls sold must be greater than or equal to zero.
The solution [tex]\( x = -6.25 \)[/tex] is not viable because the number of dolls sold cannot be a negative number.
The solution [tex]\( x = 10 \)[/tex] is viable because it represents a non-negative number of dolls sold.
Thus, out of the two total solutions, only one solution is viable.
So, the correct answers are:
Total possible solutions: 2
Viable solutions: 1
