Question

Assuming the market price of coal is p=$300 (inverse supply P=0.2Q), calculate:

A. Quantity supplied

B.MWTA by producers

C.MC of production

D. WTA by producers

E. Total variable costs

F. total revenue

G. producer surplus

Answer 1

(A) Assuming perfect competition, quantity supplied would be $\dpi{80} 300 = 0.2Q$ or $\dpi{80} Q = 1500$ units.

(B) The supply curve represents the willingness to accept (WTA), and hence, the marginal willingness to accept (MWTA) would be the amount of WTA per unit quantity. The WTA is $\dpi{80} MWTA = \frac{\partial P}{\partial Q}$ or $\dpi{80} MWTA = \frac{\partial }{\partial Q}(0.2Q)$ or $\dpi{80} MWTA = 0.2$ .

(C) The supply curve is determined as $\dpi{80} P = MC$ over the average variable cost. In this case, the supply is $\dpi{80} P = 0.2Q$ , and hence, the marginal cost will be $\dpi{80} MC = 0.2Q$ . As the MC is linear, we may say that it is never below AVC for Q>0.

(D) WTA by producers is $\dpi{80} P = 0.2Q$ . At current equilibrium quantity of 1500 units, the WTA of producers is the equilibrium price as $\dpi{80} P = 0.2Q = 0.2*1500 = 300$ .

(E) The marginal cost is $\dpi{80} MC = 0.2Q$ , and is found by $\dpi{80} MC = \frac{\partial }{\partial Q}(TC)$ , and hence we have $\dpi{80} TC = \int MC \mathrm{d}Q$ or $\dpi{80} TC = \int 0.2Q \mathrm{d}Q$ or $\dpi{80} TC = 0.2\frac{Q^2}{2} + F$ or $\dpi{80} TC = 0.1Q^2 + F$ , where F - the fixed cost, is the integral constant. The total variable cost would be hence $\dpi{80} TVC = 0.1Q^2$ (since TC is TVC+F).

(F) The total revenue would be $\dpi{80} TR = Q.P$ or $\dpi{80} TR = Q(0.2Q)$ or $\dpi{80} TR = 0.2Q^2$ . In this case, the total revenue would be $\dpi{80} TR = 0.2(1500)^2 = 450000$ dollars.

(G) The cost of producers would be the area below the supply curve as $\dpi{80} \int^{Q^*}_0 P.\mathrm{d}Q$ , where Q* is the equilibrium quantity. Hence, we have $\dpi{80} \int^{Q^*}_0 P.\mathrm{d}Q = \int^{1500}_0 0.2Q .\mathrm{d}Q$ or $\dpi{80} \int^{Q^*}_0 P.\mathrm{d}Q = 0.2 \int^{1500}_0 Q .\mathrm{d}Q$ or $\dpi{80} \int^{Q^*}_0 P.\mathrm{d}Q = 0.2 [\frac{Q^2}{2}]^{1500}_0$ or $\dpi{80} \int^{Q^*}_0 P.\mathrm{d}Q = 0.2 [\frac{1500^2 - 0^2}{2}]$ or $\dpi{80} \int^{Q^*}_0 P.\mathrm{d}Q = 0.1 [1500^2] = 225000$ . The producer surplus would be hence total revenue minus this cost, ie $\dpi{80} PS = 225000$ dollars.