Maximizing Profits When Demand is Semi-Log

We want to maximize profits, or (P-MC)*Q. Our demand function is given as ln(Q) = aP + b. To get Q from ln(Q), we use the exponential function (since e^ln(x) = x):

e^lnQ = e^(aP+b)

Plugging back into our profit function, we have:

Π =(P−MC)*Q

Π =(P−MC)* e^(aP+b)

To maximize profits, we take the derivative with respect to P and set it equal to zero.

∂/∂P(Π) = (P – MC) * a * e^(aP + b) + e^(aP + b) = (1 + aP – aMC) * e^(aP + b) = 0

(see note at end on how to calculate this derivative)

In order for the derivative to equal zero, either (1+ aP – aMC) or (eaP+b) must be zero. Since e x can never equal 0 for a real value of x, we know that it must be the first term, 1+ aP – aMC, that equals zero.

We can now solve for P: 1+aP−aMC =0 aP = aMC −1 P =MC− 1/a

Note on derivative: In order to see how the derivative is calculated, note the product rule for derivatives: (∂(g*h))/dx = (g* (∂h)/dx) + (h* ∂g/dx). In this case, g = (P-MC) and h = eaP+b. Note also that the derivate of e f(x) with respect to x is (df/dx) * e f(x).

Maximizing Profits When Demand is Linear

We want to maximize profits when Q=aP +b

Profit = (P-MC) (Q) = (P-MC) (aP+b)

Multiplying through and differentiating profit with respect to price and setting = 0 for the maxima , we get 2aP-aMC + b = 0.