Question

Given the demand function. P=16𝑒−0.02𝑄

• Determine the quantity and price at which total revenue will be maximized.

• test the second-order condition.

Answer 1

1.

$P=16e^{-0.02Q}$

Total revenue$=TR=PQ=16e^{-0.02Q}.Q$

$\frac{\delta TR}{\delta Q}=16e^{-0.02Q} +16Q(-0.02)e^{-0.02Q}=0$

$16e^{-0.02Q}[1-0.02Q]=0$

$[e^{-0.02Q }$ is always positive]

so, $1-0.02Q=0$

$Q=\frac{1}{0.02}=50$

$P=16e^{-0.02(50)}$

$p=\frac{16}{e}=\frac{16}{2.72}=5.88$

(b)

$\frac{\delta ^2TR}{\delta Q^2}=16e^{-0.02Q}(-0.02)+(-0.032)e^{-0.02Q}$

$=-0.32Qe^{-0.02}(0.02)$

<0

=negative

second order is also satisfied.