Suppose your demand function is given by D(q) = - q? - 2q + 597, where q is thousands of units sold and D(q) is dollars per unit. Compute the following, showing all calculations clearly.
A) If 9000 units are to be sold, what price should be charged for the item?
Price = $ 498
B) If a price of $477 is set for this item, how many units can you expect to sell? (Give your answer as whole
units, not in thousands of units.)
You can sell 9396
x whole units (Your answer should not be terms of thousands of units).
C) At what value of q does D(9) cross the q axis? (When you give your answer, round your answer to three
decimal places)
It crosses at q =
597
x thousand units.

Part A

q = 9 represents selling 9000 units, since q is thousands of units sold

Plug this into the D(q) function

D(q) = -q^2 - 2q + 597

D(9) = -(9)^2 - 2(9) + 597

D(9) = 498

The price per unit should be $498

You have the correct answer.

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Part B

Plug in D(q) = 477. Then solve for x.

D(q) = -q^2 - 2q + 597

477 = -q^2 - 2q + 597

-q^2 - 2q + 597 = 477

-q^2 - 2q + 597-477 = 0

-q^2 - 2q + 120 = 0

q^2 + 2q - 120 = 0

(q+12)(q-10) = 0

q+12 = 0 or q-10 = 0

q = -12 or q = 10

Ignore negative q values. It is not possible to have negative demand.

So if the unit price is $477, then you can expect to sell 10,000 units.

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Part C

Plug D(q) = 0 and solve for q

-q^2 - 2q + 597 = 0

q^2 + 2q - 597 = 0

1q^2 + 2q + (-597) = 0

1x^2 + 2x + (-597) = 0

We have an equation in the form ax^2+bx+c = 0 with a = 1, b = 2, c = -597

Use the quadratic formula to solve for x

We ignore any negative solution. The only practical solution is roughly x = 23.454, so q = 23.454 is when D(q) is equal to 0

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