To determine the price range of mini bread loaves (per dozen) that would allow PJ's bakery to earn over $1,800 in profits, we need to solve the inequality based on the profit function:
\[ p(x) = -3x^2 + 150x \]
We want to find when:
\[ -3x^2 + 150x > 1800 \]
Let's rearrange this inequality:
\[ -3x^2 + 150x - 1800 > 0 \]
Now, we can multiply the entire inequality by -1 (which flips the inequality sign):
\[ 3x^2 - 150x + 1800 < 0 \]
Next, we need to solve the quadratic equation:
\[ 3x^2 - 150x + 1800 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here:
- \( a = 3 \)
- \( b = -150 \)
- \( c = 1800 \)
First, calculate the discriminant:
\[ b^2 - 4ac = (-150)^2 - 4(3)(1800) \] \[ = 22500 - 21600 \] \[ = 900 \]
Now, applying the quadratic formula:
\[ x = \frac{-(-150) \pm \sqrt{900}}{2(3)} \] \[ = \frac{150 \pm 30}{6} \]
Calculating the two solutions:
- \( x = \frac{150 + 30}{6} = \frac{180}{6} = 30 \)
- \( x = \frac{150 - 30}{6} = \frac{120}{6} = 20 \)
Now we have the quadratic \( 3x^2 - 150x + 1800 < 0 \). The critical points are \( x = 20 \) and \( x = 30 \).
The parabola opens upwards (since the coefficient of \( x^2 \) is positive), so the profit will be above $1,800 in the intervals outside the roots:
- \( x < 20 \)
- \( x > 30 \)
Thus, the range of prices that allows PJ's bakery to earn over $1,800 in profits is either:
- Less than $20.00
- Greater than $30.00
So, the correct response is:
The bread should be priced either less than $20.00 or greater than $30.00 to make a profit of $1,800.00.
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