Question

PJ owns a small bakery where fresh bread is baked and sold daily. The profit function p(x)=−3x2+150x models the bakery’s profits, in dollars, as a function of the cost per dozen of mini bread loaves. If PJ wants to earn over $1,800.00 in profits, then what does the price range of the mini bread loaves (per dozen) need to be?(1 point)%0D%0AResponses%0D%0A%0D%0AThe bread should be priced greater than $0.00 and less than $60.00 to make a profit of $1,800.00.%0D%0AThe bread should be priced greater than $0.00 and less than $60.00 to make a profit of $1,800.00.%0D%0A%0D%0AThe bread should be priced either less than $20.00 or greater than $30.00 to make a profit of $1,800.00.%0D%0AThe bread should be priced either less than $20.00 or greater than $30.00 to make a profit of $1,800.00.%0D%0A%0D%0AThe bread should be priced between $20.00 and $30.00 to make a profit of $1,800.00.%0D%0AThe bread should be priced between $20.00 and $30.00 to make a profit of $1,800.00.%0D%0A%0D%0AThe bread should be priced either less than $4.38 or greater than $45.62 to make a profit of $1,800.00.

Answer 1

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:

1. \( x = \frac{150 + 30}{6} = \frac{180}{6} = 30 \)
2. \( x = \frac{150 - 30}{6} = \frac{120}{6} = 20 \)

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:

1. \( x < 20 \)
2. \( x > 30 \)

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.**