Question 6.38: A firm’s demand function is given by the equation P = 150/e^......

(a) The demand function is sketched in Figure 6.40 from the points in Table 6.21.

(b) TR = PQ = 150/e^{-0.02Q}

To find MR, differentiate TR. As TR is a product, use the product rule to differentiate w.r.t. Q.
Let u = 150Q and v = e^{-0.02Q}, hence,

\frac {du}{dQ}= 150    and    \frac {dv}{dQ} = −0.02e^{−0.02Q}

Now, substitute u, du/dQ, v and dv/dQ into the equation for the product rule, equation 6.32:

\frac {d(TR)}{dQ} = v \frac {du}{dQ} + u \frac {dv}{dQ}

= e^{−0.02Q}(150) + 150Q(−0.02e^{−0.02Q})

= 150e^{−0.02Q}(1 − 0.02Q)   factoring out the common terms, 150/e^{-0.02Q}.

Therefore, MR = 150e^{−0.02Q}(1 − 0.02Q).

(c) Work through the usual max/min method to find the turning points on the TR curve.

Step 1: Get derivatives,

\frac {d(TR)}{dQ} = 150e^{-0.02Q}(1 − 0.02Q)

d²(TR)/dQ² is found by differentiating d(TR)/dQ which is a product. The reader is asked to show that

\frac {d²(TR)}{dQ²} = 150e^{-0.02Q}(0.0004Q − 0.04)

Step 2: Solve \frac {d(TR)}{dQ} = 0

150e^{-0.02Q}(1 − 0.02Q)

\underbrace{ 150/e^{-0.02Q}}_{\neq 0}   \underbrace{(1 –  0.02Q)}_{1−0.02Q=0→Q=50} = 0

Step 3: Max or min?

Evaluate the second derivative at Q = 50:

TR″ = 150e^{−0.02(50)}[0.0004(50) − 0.04] = 55.182[−0.02] = −1.1036 < 0

Since TR″ < 0, TR is a maximum at Q = 50, TR = 2759.1.