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
[tex]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\\\\[/tex]
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
