Relationship Between Price and Demand

Question

A company has found that the relationship between the price p and the demand x for a particular product is given approximately byp=1252−0.18x2.The company also knows that the cost of producing the product is given by C(x)=840+382x.

Find P(x), the profit function.

P(x) = 

Now use the profit function to do the following:

(A) Find the average of the x values of all local maxima of P.

Note: If there are no local maxima, enter -1000.

Average of x values = 

(B) Find the average of the x values of all local minima of P.

Note: If there are no local minima, enter -1000.

Average of x values = 

(C) Use interval notation to indicate where P(x) is concave up.

If you have extra boxes, fill each in with an 'x'.

Concave up:

(D) Use interval notation to indicate where P(x) is concave down.

Concave down:

Answer 1

Let's start by finding the profit function \(P(x)\), and then we'll proceed to the other parts of the question.

The profit function \(P(x)\) is given by the difference between the revenue and the cost:

\[P(x) = R(x) - C(x)\]

Where:

- \(R(x)\) is the revenue, which is the product of price (\(p\)) and demand (\(x\)): \(R(x) = p \cdot x\).

- \(C(x)\) is the cost function given as \(C(x) = 840 + 382x\).

- \(p\) is the price function given as \(p = 1252 - 0.18x^2\).

So, we have:

\[R(x) = (1252 - 0.18x^2) \cdot x\]

\[C(x) = 840 + 382x\]

Now, let's calculate \(P(x)\):

\[P(x) = R(x) - C(x) = (1252 - 0.18x^2) \cdot x - (840 + 382x)\]

Simplify \(P(x)\):

\[P(x) = 1252x - 0.18x^3 - 840 - 382x\]

\[P(x) = -0.18x^3 + 870x - 840\]

Now, we have the profit function \(P(x) = -0.18x^3 + 870x - 840\).

(A) To find the average of the \(x\) values of all local maxima of \(P\), we need to find the derivative of \(P(x)\) and locate its critical points.

Taking the derivative of \(P(x)\):

\[P'(x) = -0.54x^2 + 870\]

Setting \(P'(x)\) equal to zero to find critical points:

\[-0.54x^2 + 870 = 0\]

\[-0.54x^2 = -870\]

\[x^2 = \frac{870}{0.54}\]

\[x^2 \approx 1611.11\]

\[x \approx \pm \sqrt{1611.11}\]

Since we're looking for local maxima, we'll consider the positive square root:

\[x \approx \sqrt{1611.11} \approx 40.14\]

So, we have one local maximum at \(x \approx 40.14\). Now, let's move on to finding local minima.

(B) To find the average of the \(x\) values of all local minima of \(P\), we'll continue with the critical points we've found.

We only found one critical point, which is also a local maximum. Therefore, there are no local minima. We'll enter -1000 as indicated in the question.

(C) To find where \(P(x)\) is concave up, we need to examine the second derivative of \(P(x)\) and look for intervals where it's positive.

Taking the second derivative of \(P(x)\):

\[P''(x) = -1.08x\]

Since the coefficient of \(x\) in \(P''(x)\) is negative (-1.08), \(P(x)\) is concave down for all values of \(x\).

(D) To find where \(P(x)\) is concave down, we already determined that \(P(x)\) is concave down for all values of \(x\). Therefore, the answer is all real numbers, represented as \((-\infty, \infty)\).

In interval notation:

- \(P(x)\) is concave up: None (No intervals)

- \(P(x)\) is concave down: \((-\infty, \infty)\)