Question

question 5
suppose your demand function is given by $d(q) = -q^2 - 2q + 571$, where $q$ is thousands of units sold and $d(q)$ is dollars per unit. compute the following, showing all calculations clearly.
a) if 12000 units are to be sold, what price should be charged for the item?
price = $ square $
b) if a price of $248 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 $ square $ whole units (your answer should not be terms of thousands of units).
c) at what value of $q$ does $d(q)$ cross the $q$ axis? (when you give your answer, round your answer to three decimal places)
it crosses at $q = square $ thousand units.
question help: video

Explanation:

Part A

Step1: Determine the value of \( q \)

Since \( q \) is in thousands of units, for 12000 units, \( q = \frac{12000}{1000} = 12 \).

Step2: Substitute \( q = 12 \) into the demand function

The demand function is \( D(q) = -q^2 - 2q + 571 \). Substitute \( q = 12 \):
\[

$$\begin{align*} D(12) &= -(12)^2 - 2(12) + 571 \\ &= -144 - 24 + 571 \\ &= -168 + 571 \\ &= 403 \end{align*}$$

\]

Step1: Set up the equation

We know \( D(q) = 248 \), so substitute into the demand function:
\[
248 = -q^2 - 2q + 571
\]
Rearrange to form a quadratic equation:
\[
q^2 + 2q - 323 = 0
\]

Step2: Solve the quadratic equation

Use the quadratic formula \( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 2 \), \( c = -323 \):
\[

$$\begin{align*} q &= \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-323)}}{2(1)} \\ &= \frac{-2 \pm \sqrt{4 + 1292}}{2} \\ &= \frac{-2 \pm \sqrt{1296}}{2} \\ &= \frac{-2 \pm 36}{2} \end{align*}$$

\]
We have two solutions:
\[
q_1 = \frac{-2 + 36}{2} = \frac{34}{2} = 17, \quad q_2 = \frac{-2 - 36}{2} = \frac{-38}{2} = -19
\]
Since \( q \) (thousands of units) cannot be negative, \( q = 17 \). To find the number of whole units, multiply by 1000: \( 17 \times 1000 = 17000 \).

Step1: Set \( D(q) = 0 \)

To find where \( D(q) \) crosses the \( q \)-axis, set \( D(q) = 0 \):
\[
-q^2 - 2q + 571 = 0
\]
Multiply both sides by -1:
\[
q^2 + 2q - 571 = 0
\]

Step2: Solve the quadratic equation

Use the quadratic formula \( q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 2 \), \( c = -571 \):
\[

$$\begin{align*} q &= \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-571)}}{2(1)} \\ &= \frac{-2 \pm \sqrt{4 + 2284}}{2} \\ &= \frac{-2 \pm \sqrt{2288}}{2} \\ &= \frac{-2 \pm 47.833}{2} \end{align*}$$

\]
We have two solutions:
\[
q_1 = \frac{-2 + 47.833}{2} \approx \frac{45.833}{2} \approx 22.916, \quad q_2 = \frac{-2 - 47.833}{2} \approx \frac{-49.833}{2} \approx -24.916
\]
Since \( q \) cannot be negative, we take \( q \approx 22.916 \).

Answer:

\( 403 \)

Part B