The problem gives the demand function $p = 100e^{-q/2}$, where $p$ is the price per unit and $q$ is the number of units. (a) We need to find the price $p$ when the quantity demanded $q = 6$ units. The answer should be rounded to the nearest cent. (b) We need to find the quantity demanded $q$ when the price $p = \$2.01$ per unit. The answer should be rounded to the nearest unit.

1. Problem Description

The problem gives the demand function $p = 100e^{-q/2}$, where $p$ is the price per unit and $q$ is the number of units.

(a) We need to find the price $p$ when the quantity demanded $q = 6$ units. The answer should be rounded to the nearest cent.

(b) We need to find the quantity demanded $q$ when the price p = \2.01$ per unit. The answer should be rounded to the nearest unit.

2. Solution Steps

(a)

We are given $q = 6$. Substituting this into the demand function $p = 100e^{-q/2}$, we have

$p = 100e^{-6/2} = 100e^{-3}$.

$p \approx 100(0.049787) \approx 4.9787$.

Rounding to the nearest cent, we get $p = 4.98$.

(b)

We are given $p = 2.01$. Substituting this into the demand function $p = 100e^{-q/2}$, we have

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

Divide both sides by 100:

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

Take the natural logarithm of both sides:

$\ln(0.0201) = \ln(e^{-q/2})$

$\ln(0.0201) = -q/2$

$q = -2\ln(0.0201)$

$q \approx -2(-3.9084) \approx 7.8168$.

Rounding to the nearest unit, we get $q = 8$.

3. Final Answer

(a)

$4.98$

(b)

$8$