Answer:
(a) the price per units is $5.0
(b)the number of units demanded = 7
Step-by-step explanation:
The demand function is given by ![p=100e^{-q/2}](https://tex.z-dn.net/?f=p%3D100e%5E%7B-q%2F2%7D)
(a) Now, we have to find the value of p when q = 6
Substitute q = 6 in the above equation
![p=100e^{-6/2}\\\p=100e^{-3}](https://tex.z-dn.net/?f=p%3D100e%5E%7B-6%2F2%7D%5C%5C%5Cp%3D100e%5E%7B-3%7D)
On simplifying, we get
![p=4.97870683679](https://tex.z-dn.net/?f=p%3D4.97870683679)
Rounding to nearest cents, we have
![p=5.0](https://tex.z-dn.net/?f=p%3D5.0)
Therefore, the price per units is $5.0
(b)
Now, we have to find q for p = $2.99
![2.99=100e^{-q/2}](https://tex.z-dn.net/?f=2.99%3D100e%5E%7B-q%2F2%7D)
Divide both sides by 100
![0.0299=e^{-q/2}](https://tex.z-dn.net/?f=0.0299%3De%5E%7B-q%2F2%7D)
Take natural log both sides
![\ln(0.0299)=\ln(e^{-q/2})](https://tex.z-dn.net/?f=%5Cln%280.0299%29%3D%5Cln%28e%5E%7B-q%2F2%7D%29)
On simplifying, we get
![\ln(0.0299)=-q/2\ln(e)\\\\\ln(0.0299)=-q/2\\\\q=-2\ln \left(0.0299\right)\\\\q=7.0](https://tex.z-dn.net/?f=%5Cln%280.0299%29%3D-q%2F2%5Cln%28e%29%5C%5C%5C%5C%5Cln%280.0299%29%3D-q%2F2%5C%5C%5C%5Cq%3D-2%5Cln%20%5Cleft%280.0299%5Cright%29%5C%5C%5C%5Cq%3D7.0)
Therefore, the number of units demanded = 7