To find the price at which Bella should sell her book to maximize her revenue, we need to analyze the equation given:
\[ y = -14x^2 + 6x \]
This is a quadratic equation in the standard form \( y = ax^2 + bx + c \), where:
- \( a = -14 \) (which is less than 0, indicating a parabola that opens downwards)
- \( b = 6 \)
- \( c = 0 \) (we can ignore \( c \) for this context since we are only interested in the revenue-maximizing price)
The maximum value of a quadratic equation occurs at the vertex, which can be found using the formula:
\[ x = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ x = -\frac{6}{2(-14)} \] \[ x = -\frac{6}{-28} \] \[ x = \frac{6}{28} \] \[ x = \frac{3}{14} \]
However, since this is unrealistic for a price in this context, let's calculate using a more complete method involving the product of the format \( y = -14x^2 + 6x \).
We'll plot the value of \( x \) using the direct increments from the choices provided:
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Checking prices \( x = 6 \): \[ y = -14(6)^2 + 6(6) = -14(36) + 36 = -504 + 36 = -468 \]
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Checking prices \( x = 12 \): \[ y = -14(12)^2 + 6(12) = -14(144) + 72 = -2016 + 72 = -1944 \]
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Checking prices \( x = 24 \): \[ y = -14(24)^2 + 6(24) = -14(576) + 144 = -8064 + 144 = -7920 \]
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Checking prices \( x = 36 \): \[ y = -14(36)^2 + 6(36) = -14(1296) + 216 = -18144 + 216 = -17928 \]
After analyzing the values, we can see:
The quadratic opens downward, the maximum occurs at \( x = \frac{3}{14} \), and hasn't shown realistic context vs options provided.
Given common pricing, and if bounding where trials showcase no positive income with standard checks for hypotheses and curve fitting, the maximum price to charge could be \( 6 \), as least distortion of loss revenue noted hence.
Thus, the price should be $6 to earn near maximum revenue, reaffirming context for realistic pricing to attempt actualized returns without redemption hashing faults.