Question

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 = $ 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 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 = $ thousand units.

Explanation:

Part A

Step1: Determine q value

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

Step2: Substitute q into D(q)

Substitute \( q = 12 \) into \( D(q)=-q^{2}-2q + 571 \).
\[

$$\begin{align*} D(12)&=- (12)^{2}-2\times12 + 571\\ &=-144-24 + 571\\ &=-168 + 571\\ &=403 \end{align*}$$

\]

Step1: Set up the equation

Set \( D(q)=248 \), so we have the equation \( -q^{2}-2q + 571=248 \).

Step2: Rearrange the equation

Rearrange it to standard quadratic form \( -q^{2}-2q+571 - 248 = 0\), which simplifies to \( -q^{2}-2q + 323 = 0\). Multiply both sides by - 1 to get \( q^{2}+2q - 323=0 \).

Step3: Solve the quadratic equation

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

$$\begin{align*} q&=\frac{-2\pm\sqrt{2^{2}-4\times1\times(-323)}}{2\times1}\\ &=\frac{-2\pm\sqrt{4 + 1292}}{2}\\ &=\frac{-2\pm\sqrt{1296}}{2}\\ &=\frac{-2\pm36}{2} \end{align*}$$

\]
We have two solutions: \( q=\frac{-2 + 36}{2}=\frac{34}{2}=17 \) and \( q=\frac{-2-36}{2}=\frac{-38}{2}=-19 \). Since \( q \) (thousands of units) can't be negative, we take \( q = 17 \). The number of units is \( 17\times1000 = 17000 \).

Step1: Set D(q) = 0

We need to solve \( -q^{2}-2q + 571=0 \). Multiply both sides by - 1 to get \( q^{2}+2q - 571=0 \).

Step2: Apply quadratic formula

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

$$\begin{align*} q&=\frac{-2\pm\sqrt{2^{2}-4\times1\times(-571)}}{2\times1}\\ &=\frac{-2\pm\sqrt{4 + 2284}}{2}\\ &=\frac{-2\pm\sqrt{2288}}{2}\\ &=\frac{-2\pm47.833}{2} \end{align*}$$

\]
We have two solutions: \( q=\frac{-2 + 47.833}{2}=\frac{45.833}{2}=22.9165\approx22.917 \) and \( q=\frac{-2-47.833}{2}=\frac{-49.833}{2}=-24.9165 \). Since \( q \) (thousands of units) can't be negative, we take the positive solution.

Answer:

\( 403 \)

Part B