Simpsons, inc. can sell 500 pairs of boots per week if they charge $100 sales drop to 480 pairs per week if they charge $108. assume the price p is a linear function of the weekly demand for boots, q.
complete parts a-d below.

(a) complete the table:

demand, q price $
460
480 108
500 100
520
(b) write a formula for the demand function p = d(q)
type an expression using q as the variable.
(c) find the axis intercepts of d(q). interpret the intercepts by completing the following:
1. the quantity demanded is 0 boots when the price for a pair of boots is $
2. at a price of $0 (free), the demand is pairs of boots.
(d) interpret the slope of the line by completing the following:
1. lowering the price by s24 results in more pairs of boots sold per week.
2. raising the price by $ results in 70 fewer pairs sold per week.

  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 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).

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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

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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.

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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.

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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)

it would more likely be bald eagle because its has more in the box so about 75%.

step-by-step explanation: