The problem provides a demand function $p = 100e^{-q/2}$, where $p$ is the price per unit and $q$ is the number of units. Part (a) asks to find the price $p$ when the quantity demanded is $q=6$ units. Part (b) asks to find the quantity demanded $q$ when the price is $p=2.01$ per unit.

1. Problem Description

The problem provides a demand function $p = 100e^{-q/2}$, where $p$ is the price per unit and $q$ is the number of units. Part (a) asks to find the price $p$ when the quantity demanded is $q=6$ units. Part (b) asks to find the quantity demanded $q$ when the price is $p=2.01$ per unit.

2. Solution Steps

(a) We are given $q=6$. We need to find $p$.

Substitute $q=6$ into the demand function:

$p = 100e^{-6/2}$

$p = 100e^{-3}$

$p = 100(0.049787)$

$p = 4.9787$

Round to the nearest cent: $p = 4.98$.

(b) We are given $p=2.01$. We need to find $q$.

Substitute $p=2.01$ into the demand function:

$2.01 = 100e^{-q/2}$

Divide both sides by 100:

$0.0201 = e^{-q/2}$

Take the natural logarithm of both sides:

$ln(0.0201) = -q/2$

$-3.9082 = -q/2$

Multiply both sides by -2:

$q = 2(3.9082)$

$q = 7.8164$

Round to the nearest unit: $q = 8$.

3. Final Answer

(a) $4.98

(b) 8