Lecture Notes in Ma 20 (Calculus): Sample Word Problem in Economics Involving the Derivative, 6/30/08

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Lecture Notes in Ma 20 (Calculus for Economics)
30 June 2008

Topic:

Sample Word Problem in Economics Involving the Derivative


Suppose that the demand set for cans of salmon is

D = {(q,p) where (p^3)q = 8000} ,

where q is the number of cans (in thousands per week) and p is the price per can in dollars.


a). If p is increased from $20 to $21, what will be the expected fall in sales, approximately?

b). If, on the other hand, production were to be increased from 1000 tins per week to 1100, what would be the expected fall in price?


SOLUTION:

The demand function q and its derivative q’ are

q(p) = 8000p^-3

q’(p) =(8000)(-3)p^-4 = -24000/p^4

a). Therefore when p = 20 and (delta)p = 1 we have

(delta)q is approximately equal to (-24000)/p^4)(delta)p = -24009/20^4 = -0.15

Remembering that the units are thousands of cans, it follows that 150 fewer cans will be sold per week.

b). In letter b, we have to consider p as a function of q, and so we need the inverse demand function and its derivative:

(p^3)q = 8000

p^3 = 8000/q

p = cuberoot of (8000/q)

p = 20q^(-1/3)

p’(q) = (-1/3)(20)q^(-4/3) = (-20/3)q^(-4/3)

So when q = 1 and (delta)q = 0.1 we have

(delta)p is approximately equal to [(-20/3)q^(-4/3)](0.1) = -2/3 or approx. 0.67

The conclusion is that the price falls by about 67 cents.

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Reference: Martin Anthony and Norman Biggs (1996). Mathematics for Eonomics and Finance. Cambridge University Press, pp. 57 - 58.
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