Answer:
a) (demand, price) = (460, 116), (520, 92)
b) p = 300 -0.4q
c) $300; 750 pairs
d) 60 pairs; $28
Step-by-step explanation:
a) For each change in demand by 20 pairs, the corresponding price change is $8 in the opposite direction. 460 pairs is 20 <em>less</em> than 480 pairs, so the corresponding price is $8 <em>more</em> than the $108 price for 480 pairs. That is, (q, p) = (460, $116).
Similarly, 520 is 20 <em>more</em> pairs than the 500 sold for $100, so the corresponding price is $8 <em>less</em>. 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)
