View an example | All parts showing and S(x) is X it, (b) the co D(x) is the price, in dollars per unit, that consumers are willing to pay for x units of an item, and S(x) is the price, in dollars per unit, that producers are willing to accept for x units. Find (a) the equilibrium point, (b) the consumer surplus at the equilibrium point, and (c) the producer surplus at the equilibrium point. 196 D(x) = - = S(x) = Vx a) The equilibrium point is the intersection of the curves D(x) and S(x). Set D(x) equal to S(x) and solve for x. Set D(x) equal to S(x) and simplify. D(x) = S(x) 196 X VX 196 = X Multiply both sides by vx Now substitute for x in either D(x) or S(x) and calculate the y-coordinate of the point of intersection. This example uses S(x). S(x) = x S(196) = 196 Substitute x = 196. S(196) = 14 Simplify. The equilibrium point is (196,14). (b) The consumer surplus is defined for the point (Q.P) as follows. Jo D( x)dx - Qp Determine the consumer surplus for the equilibrium point (196,14). Substitute and integrate. [ D(x)ax- Qp = J. x x- (196)(14 ) Substitute. = [392x ] -(196)(14) Integrate. Evaluate. [392 x ]0 -(196)(14) = 392,196 -(196)(14) = 2744 The consumer surplus for the equilibrium point (196,14) is $2744. (c) The producer surplus is defined for the point (Q.P) as follows. Determine the producer surplus for the equilibrium point (196,14). Substitute and integrate. op - J. s(x)dx = (196) (14) - J ( vx ) ax Substitute. = (196)(14) - 2(x)3/2 196 Integrate. Evaluate. (196)(14) - 2(X)3/2 = (196)(14) - 2(196)3/2 ~ 914.67 The producer surplus for the equilibrium point (196,14) is $914.67.Dix} is the price, in dollars per unit. that consumers are willing to payr for x units of an item, and 30:} is the price. in dollars per unit. that producers are willing to accept for x units. Find (a) the equilibrium point. {b} the consumer surplus at the equilibrium point; and (c) the producer surplus at the equilibrium point. 1E9 nix) = 3.5% J