Question

Suppose your demand function is given by D(q) = - q? - 2q + 597, where q is thousands of units sold and

Answer 1

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

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(2)\pm\sqrt{(2)^2-4(1)(-597)}}{2(1)}\\\\x = \frac{-2\pm\sqrt{2392}}{2}\\\\x \approx \frac{-2\pm48.90807704}{2}\\\\x \approx \frac{-2+48.90807704}{2} \text{ or } x \approx \frac{-2-48.90807704}{2}\\\\x \approx \frac{46.90807704}{2} \text{ or } x \approx \frac{-50.90807704}{2}\\\\x \approx 23.45403852 \text{ or } x \approx -25.45403852$

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