Answer:
(a) The price per unit is $1.83 when the quantity demanded is 8 units
(b) The quantity demanded is approximately 8 units when the price per unit is $1.87
Explanation:
Given
[tex]p = 100e^{-q/2}[/tex]
[tex]p \to[/tex] price per unit
[tex]q \to[/tex] quantity demanded
Solving (a): Price per unit when quantity is 8
This means that we calculate p(8)
We have:
[tex]p(q) = 100e^{-q/2}[/tex]
So:
[tex]p(8) = 100e^{-8/2}[/tex]
[tex]p(8) = 100e^{-4}[/tex]
[tex]p(8) = 1.83[/tex]
Solving (b): Quantity demanded when price per unit is $1.87
This means that:
[tex]p(q) = 1.87[/tex] ---- find q
We have:
[tex]p(q) = 100e^{-q/2}[/tex]
So:
[tex]1.87 = 100e^{-q/2}[/tex]
Divide both sides by 100
[tex]0.0187 = e^{-q/2}[/tex]
Take natural logarithm of both sides
[tex]\ln(0.0187) = \ln(e^{-q/2})[/tex]
[tex]-3.980 = \ln(e^{-q/2})[/tex]
Rewrite as:
[tex]-3.980 = -q/2}*\ln(e)[/tex]
[tex]-3.980 = -q/2[/tex]
Multiply by -2
[tex]7.96 = q[/tex]
[tex]q = 7.96[/tex]
Approximate
[tex]q = 8[/tex]