Q.(a) The demand and supply function of a commodity are $P_d = 18 - 2x - x^2$ and $P_s = 2x - 3$. Find the consumer's surplus and producer's surplus at equilibrium price.

(a) $P_d=P_s\Rightarrow x_0=3,P_0=3$; $CS=\int_0^3 P_d\,dx-x_0P_0=27$, $PS=x_0P_0-\int_0^3 P_s\,dx=9$. (b) $E(X)=\tfrac45$, $V(X)=\tfrac23-\tfrac{16}{25}=\tfrac{2}{75}$.

Part (a) — Consumer's and Producer's surplus

$P_d=18-2x-x^{2}$, $P_s=2x-3$.

Step 1 — Equilibrium ($P_d=P_s$):

$18-2x-x^{2}=2x-3\ \Rightarrow\ x^{2}+4x-21=0\ \Rightarrow\ (x+7)(x-3)=0.$

Taking $x_0=3$ (positive). Then $P_0=2(3)-3=3$.

Step 2 — Consumer's surplus:

$CS=\int_{0}^{x_0}P_d\,dx-x_0P_0=\int_{0}^{3}(18-2x-x^{2})\,dx-3(3).$

$\int_{0}^{3}(18-2x-x^{2})\,dx=\left[18x-x^{2}-\frac{x^{3}}{3}\right]_{0}^{3}=54-9-9=36.$

$CS=36-9=27.$

Step 3 — Producer's surplus:

$PS=x_0P_0-\int_{0}^{x_0}P_s\,dx=9-\int_{0}^{3}(2x-3)\,dx.$

$\int_{0}^{3}(2x-3)\,dx=\left[x^{2}-3x\right]_{0}^{3}=9-9=0\ \Rightarrow\ PS=9-0=9.$

Part (b) — Expectation and variance of a continuous random variable

$f(x)=4x^{3},\ 0<x<1$.

Step 1 — $E(X)$:

$E(X)=\int_{0}^{1}x\,f(x)\,dx=\int_{0}^{1}4x^{4}\,dx=\left[\frac{4x^{5}}{5}\right]_{0}^{1}=\frac{4}{5}=0.8.$

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