Math Problem Statement

The marketing research department of a computer company used a large city to test market the firm's new laptop. The department found the relationship between price p (dollars per unit) and the demand x (units per week) was given by p = 1764 - 0.12x^2, for 0 < x < 95. Using this equation, the weekly revenue function becomes R(x) = 1764x - 0.12x^3. Find the local minimum and determine the interval(s) where the revenue function is concave upward.

Solution

Let's go step by step to address the questions from your image.

The given revenue function is:

R(x)=xp=1764x0.12x3R(x) = x \cdot p = 1764x - 0.12x^3

To solve parts A and B, we'll need to calculate the first and second derivatives of the revenue function, as these are used to determine local minima and concavity.

Step 1: First derivative (R'(x))

The first derivative helps to find the critical points:

R(x)=ddx(1764x0.12x3)=17643(0.12)x2R'(x) = \frac{d}{dx} \left(1764x - 0.12x^3 \right) = 1764 - 3(0.12)x^2

Simplify:

R(x)=17640.36x2R'(x) = 1764 - 0.36x^2

Set the first derivative equal to zero to find the critical points:

17640.36x2=01764 - 0.36x^2 = 0

Solving for xx:

0.36x2=1764x2=17640.36=49000.36x^2 = 1764 \quad \Rightarrow \quad x^2 = \frac{1764}{0.36} = 4900 x=4900=70x = \sqrt{4900} = 70

So, x=70x = 70 is a critical point. This is where the function may have a local minimum or maximum.

Step 2: Second derivative (R''(x))

Now, we calculate the second derivative to find where the graph is concave up or concave down:

R(x)=ddx(17640.36x2)=0.72xR''(x) = \frac{d}{dx}(1764 - 0.36x^2) = -0.72x

To find concavity, we analyze the sign of R(x)R''(x):

  • If R(x)>0R''(x) > 0, the graph is concave upward.
  • If R(x)<0R''(x) < 0, the graph is concave downward.

At x=70x = 70, R(70)=0.72(70)=50.4R''(70) = -0.72(70) = -50.4, which is negative, indicating concave downward at x=70x = 70.

To determine when the graph is concave upward, solve R(x)>0R''(x) > 0:

0.72x>0x<0-0.72x > 0 \quad \Rightarrow \quad x < 0

However, since xx must be between 0 and 95 (given the problem's constraints), there is no interval where the graph is concave upward.

Conclusion:

  • Part A: The local minimum occurs at x=70x = 70.
  • Part B: The graph is never concave upward.

Would you like further clarification on any of these steps?


Here are five related questions:

  1. How do you confirm if x=70x = 70 is a local minimum or maximum?
  2. What is the significance of the second derivative in determining concavity?
  3. Can a revenue function like this have multiple critical points?
  4. How would the solution change if the revenue function had a different cubic coefficient?
  5. Why is concavity important in understanding revenue models?

Tip: Always check the signs of the first and second derivatives to understand the behavior of functions, especially when dealing with maxima, minima, and concavity.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Revenue function
Critical points
Concavity
Derivatives

Formulas

First derivative: R'(x) = d/dx (1764x - 0.12x^3) = 1764 - 0.36x^2
Second derivative: R''(x) = d/dx (1764 - 0.36x^2) = -0.72x

Theorems

First Derivative Test
Second Derivative Test

Suitable Grade Level

Grades 11-12, Calculus