Answer is: option2

increasing at $0.10 per day

Solution:

To solve this problem, we need to find the rate at which the price \( p \) is changing with respect to time when the supply \( x \) is 100 crates and the supply is decreasing at a rate of \( \frac{dx}{dt} = -8 \) crates per day.

Differentiate both sides with respect to time \( t \) using the product rule: \[ \frac{d}{dt} (px) - \frac{d}{dt} (20p) - \frac{d}{dt} (6x) + \frac{d}{dt} (40) = 0 \] Using the product rule: \[ p \frac{dx}{dt} + x \frac{dp}{dt} - 20 \frac{dp}{dt} - 6 \frac{dx}{dt} = 0 \] Rearrange: \[ x \frac{dp}{dt} - 20 \frac{dp}{dt} = -p \frac{dx}{dt} + 6 \frac{dx}{dt} \] Factor: \[ (x - 20) \frac{dp}{dt} = (6 - p) \frac{dx}{dt} \] Substituting the given values:

• \( x = 100 \)
• \( \frac{dx}{dt} = -8 \)

First, solve for \( p \) by substituting \( x = 100 \) in the supply equation: \[ 100p - 20p - 6(100) + 40 = 0 \] \[ 80p - 600 + 40 = 0 \] \[ 80p = 560 \] \[ p = 7 \] Now substitute \( p = 7 \), \( x = 100 \), and \( \frac{dx}{dt} = -8 \) into the differentiated equation: \[ (100 - 20) \frac{dp}{dt} = (6 - 7)(-8) \] \[ 80 \frac{dp}{dt} = (-1)(-8) \] \[ 80 \frac{dp}{dt} = 8 \] \[ \frac{dp}{dt} = \frac{8}{80} = 0.10 \]

Since \( \frac{dp}{dt} \) is positive, the price is increasing at a rate of $0.10 per day.

The correct answer is: increasing at $0.10 per day