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Sagot :
To find the production levels that keep the production costs under [tex]$225,000$[/tex], we need to solve the inequality:
[tex]\[ C(x) = -0.52x^2 + 23x + 92 < 225 \][/tex]
First, we rearrange this inequality into a standard quadratic form:
[tex]\[ -0.52x^2 + 23x + 92 - 225 < 0 \][/tex]
Simplify the constant term:
[tex]\[ -0.52x^2 + 23x - 133 < 0 \][/tex]
Next, we need to find the roots of the quadratic equation:
[tex]\[ -0.52x^2 + 23x - 133 = 0 \][/tex]
We can use the quadratic formula, which is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our quadratic equation [tex]\( -0.52x^2 + 23x - 133 = 0 \)[/tex], the coefficients are:
- [tex]\(a = -0.52\)[/tex]
- [tex]\(b = 23\)[/tex]
- [tex]\(c = -133\)[/tex]
Using the quadratic formula, we find:
[tex]\[ x = \frac{-23 \pm \sqrt{23^2 - 4(-0.52)(-133)}}{2(-0.52)} \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = 23^2 - 4 \cdot (-0.52) \cdot (-133) \][/tex]
[tex]\[ \Delta = 529 - 2.08 \cdot 133 \][/tex]
[tex]\[ \Delta = 529 - 276.64 \][/tex]
[tex]\[ \Delta = 252.36 \][/tex]
Now plug this back into the quadratic formula:
[tex]\[ x = \frac{-23 \pm \sqrt{252.36}}{-1.04} \][/tex]
[tex]\[ x = \frac{-23 \pm 15.88}{-1.04} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{-23 + 15.88}{-1.04} \][/tex]
[tex]\[ x_1 = \frac{-7.12}{-1.04} \][/tex]
[tex]\[ x_1 \approx 6.84 \][/tex]
[tex]\[ x_2 = \frac{-23 - 15.88}{-1.04} \][/tex]
[tex]\[ x_2 = \frac{-38.88}{-1.04} \][/tex]
[tex]\[ x_2 \approx 37.38 \][/tex]
The roots of the quadratic equation are approximately [tex]\( x = 6.84 \)[/tex] and [tex]\( x = 37.38 \)[/tex].
Since the quadratic equation opens downward (the coefficient of [tex]\( x^2 \)[/tex] is negative), the inequality [tex]\( -0.52x^2 + 23x - 133 < 0 \)[/tex] is satisfied between the roots. Therefore,
[tex]\[ 6.84 < x < 37.38 \][/tex]
So, the constraint that keeps production costs under \$225,000 is [tex]\( 6.84 < x < 37.38 \)[/tex].
None of the options exactly match this solution, but the closest in terms of the valid intervals given would be:
[tex]\[ 0 \leq x < 6.84 \quad \text{and} \quad 37.39 < x \leq 47.92 \][/tex]
This option includes intervals that exclude the costly production between [tex]\( 6.84 \)[/tex] and [tex]\( 37.38 \)[/tex].
[tex]\[ C(x) = -0.52x^2 + 23x + 92 < 225 \][/tex]
First, we rearrange this inequality into a standard quadratic form:
[tex]\[ -0.52x^2 + 23x + 92 - 225 < 0 \][/tex]
Simplify the constant term:
[tex]\[ -0.52x^2 + 23x - 133 < 0 \][/tex]
Next, we need to find the roots of the quadratic equation:
[tex]\[ -0.52x^2 + 23x - 133 = 0 \][/tex]
We can use the quadratic formula, which is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our quadratic equation [tex]\( -0.52x^2 + 23x - 133 = 0 \)[/tex], the coefficients are:
- [tex]\(a = -0.52\)[/tex]
- [tex]\(b = 23\)[/tex]
- [tex]\(c = -133\)[/tex]
Using the quadratic formula, we find:
[tex]\[ x = \frac{-23 \pm \sqrt{23^2 - 4(-0.52)(-133)}}{2(-0.52)} \][/tex]
Calculate the discriminant:
[tex]\[ \Delta = 23^2 - 4 \cdot (-0.52) \cdot (-133) \][/tex]
[tex]\[ \Delta = 529 - 2.08 \cdot 133 \][/tex]
[tex]\[ \Delta = 529 - 276.64 \][/tex]
[tex]\[ \Delta = 252.36 \][/tex]
Now plug this back into the quadratic formula:
[tex]\[ x = \frac{-23 \pm \sqrt{252.36}}{-1.04} \][/tex]
[tex]\[ x = \frac{-23 \pm 15.88}{-1.04} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{-23 + 15.88}{-1.04} \][/tex]
[tex]\[ x_1 = \frac{-7.12}{-1.04} \][/tex]
[tex]\[ x_1 \approx 6.84 \][/tex]
[tex]\[ x_2 = \frac{-23 - 15.88}{-1.04} \][/tex]
[tex]\[ x_2 = \frac{-38.88}{-1.04} \][/tex]
[tex]\[ x_2 \approx 37.38 \][/tex]
The roots of the quadratic equation are approximately [tex]\( x = 6.84 \)[/tex] and [tex]\( x = 37.38 \)[/tex].
Since the quadratic equation opens downward (the coefficient of [tex]\( x^2 \)[/tex] is negative), the inequality [tex]\( -0.52x^2 + 23x - 133 < 0 \)[/tex] is satisfied between the roots. Therefore,
[tex]\[ 6.84 < x < 37.38 \][/tex]
So, the constraint that keeps production costs under \$225,000 is [tex]\( 6.84 < x < 37.38 \)[/tex].
None of the options exactly match this solution, but the closest in terms of the valid intervals given would be:
[tex]\[ 0 \leq x < 6.84 \quad \text{and} \quad 37.39 < x \leq 47.92 \][/tex]
This option includes intervals that exclude the costly production between [tex]\( 6.84 \)[/tex] and [tex]\( 37.38 \)[/tex].
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