a) For each change in demand by 20 pairs, the corresponding price change is $8 in the opposite direction. 460 pairs is 20 less than 480 pairs, so the corresponding price is $8 more than the $108 price for 480 pairs. That is, (q, p) = (460, $116).

Similarly, 520 is 20 more pairs than the 500 sold for $100, so the corresponding price is $8 less. That is, (q, p) = (520, $92).

b) The two-point form of the equation of a line will do.

p = (p2 -p1)/(q2 -q1)(q -q1) +p1

p = (100 -108)/(500 -480)(q -480) +108 . . . . substitute given table values

p = -8/20(q -480) +108

p = -0.4q +300 . . . . . . . simplify

c1) When q=0, p = -0.4·0 +300 = 300

A price of $300 will bring demand to zero.

c2) When p=0, q can be found from 0 = -0.4q +300.

q = 300/0.4 = 750

When boots are free, 750 pairs will be demanded.

d1) The derivative of price with respect to quantity is the coefficient of q in the equation:

dp/dq = -0.4

Then ...

dq/dp = 1/(-0.4) = -2.5

For a change in price of -$24, the change in quantity sold will be ...

Δq = (dq/dp)·Δp = (-2.5)(-24) = 60

60 more pairs will be sold if the price is reduced by $24.

d2) As for the previous question, ...

Δp = (dp/dq)·Δq = (-0.4)(-70) = 28

Raising the price by $28 will result in 70 fewer pairs sold.

Using derivatives is a fancy way to say the relationship between change in price and change in quantity is ...

Δq : Δp = 20 : -8 (given in the problem)

= 60 : -24 (part d1 -- multiply by 3)

= -70 : 28 (part d2 -- multiply by -3.5)

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2 years ago