a) In perfect competition, demand and supply function of commodity are given by P d = 40 x^2 and P s = 3x^2 + 8x + 8 . Discuss the nature of consumer surplus and producer surplus at the market…

a) Consumer surplus and producer surplus at equilibrium

Step 1 – Find equilibrium quantity and price

$$\begin{array}{rl}{P}_d& =40-x^2\\ P_s& =3x^2+8x+8\\ \text{Equilibrium: }& P_d=P_s\\ 40-x^2& =3x^2+8x+8\\ 4x^2+8x-32& =0\\ x^2+2x-8& =0\\ (x+4)(x-2)=0\Rightarrow x=2(\text{positive quantity})& \end{array}$$

$$P^{\ast }=P_d(2)=40-2^2=36$$

Step 2 – Consumer surplus (CS)

$$\text{CS}={\int }_0^2[P_d(x)-P^{\ast }]dx={\int }_0^2[(40-x^2)-36]dx={\int }_0^2(4-x^2)dx$$

$$\begin{array}{rl}\text{CS}& =[4x-\frac{x^3}{3}{]}_0^2=(8-\frac{8}{3})-0=\frac{16}{3}\text{units}\end{array}$$

Step 3 – Producer surplus (PS)

$$\text{PS}={\int }_0^2[P^{\ast }-P_s(x)]dx={\int }_0^2[36-(3x^2+8x+8)]dx={\int }_0^2(28-8x-3x^2)dx$$

$$\begin{array}{rl}\text{PS}& =[28x-4x^2-x^3{]}_0^2=(56-16-8)-0=32\text{units}\end{array}$$

Interpretation

The graph below visualises the two curves, the equilibrium point $(Q^{\ast }=2,P^{\ast }=36)$ and the shaded surplus areas.

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b) Homogeneous first‑order, first‑degree differential equation

A typical homogeneous equation of this class is

$$\frac{dy}{dx}=\frac{x+y}{x-y}$$

Because the right‑hand side is a function of the ratio $y/x$ only, the equation is homogeneous.

Solution

1. Set $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$.

2. Substitute:

$$v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}$$

3. Isolate $dv/dx$:

$$x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}=\frac{1+v^2}{1-v}$$

$$\frac{dv}{dx}=\frac{1+v^2}{x(1-v)}$$

4. Separate variables:

$$\frac{1-v}{1+v^2}dv=\frac{dx}{x}$$

5. Integrate:

$$\int [\frac{1}{1+v^2}-\frac{v}{1+v^2}]dv=\int \frac{dx}{x}$$

$$\arctan v-\frac{1}{2}\ln (1+v^2)=\ln |x|+C$$

6. Replace $v=y/x$:

$$\boxed{\arctan \left(\frac{y}{x}\right)-\frac{1}{2}\ln (1+(\frac{y}{x}{)}^2)=\ln |x|+C}$$

This implicit relation gives the general solution of the homogeneous first‑order, first‑degree differential equation.

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c) Relationship between price elasticity, demand and total revenue

The price elasticity of demand ($\epsilon$) is defined as

$$\epsilon =\frac{\%\Delta Q_d}{\%\Delta P}=\frac{d{Q}_d}{dP}\frac{P}{Q_d}$$

Total revenue (TR) is the product of price and quantity demanded:

$$TR=P{Q}_d(P)$$

Differentiating TR with respect to price:

$$\frac{d(TR)}{dP}=Q_d+P\frac{d{Q}_d}{dP}=Q_d[1+\frac{P}{Q_d}\frac{d{Q}_d}{dP}]=Q_d[1+\epsilon ]$$

Implications

Thus, the magnitude of price elasticity directly determines whether a change in price will increase or decrease total revenue, linking the concepts of demand responsiveness and revenue generation.