Math Problem Statement

A company selling widgets has found that the number of items sold x depends upon the price p at which they're sold, according to the equation x = 70000 / sqrt(2p + 1). Due to inflation and increasing health benefit costs, the company has been increasing the price by $4 per month. Find the rate at which revenue is changing when the company is selling widgets at $150 each.

Solution

The problem asks for the rate at which revenue is changing with respect to time when the price of the widgets is increasing by $4 per month and the current price is $150.

We are given the number of items sold xx in terms of the price pp by the equation:

x=70,0002p+1x = \frac{70,000}{\sqrt{2p + 1}}

Revenue RR is the product of the number of items sold and the price per item:

R=xp=70,0002p+1pR = x \cdot p = \frac{70,000}{\sqrt{2p + 1}} \cdot p

To find the rate at which revenue is changing when p=150p = 150, we need to calculate dRdt\frac{dR}{dt}, where dpdt=4dollars/month\frac{dp}{dt} = 4 \, \text{dollars/month} (the price is increasing by $4 per month).

Steps:

  1. First, differentiate the revenue function RR with respect to pp using the product rule: dRdp=ddp(70,000p2p+1)\frac{dR}{dp} = \frac{d}{dp} \left( \frac{70,000 \cdot p}{\sqrt{2p + 1}} \right)

  2. Then use the chain rule to find dRdt=dRdpdpdt\frac{dR}{dt} = \frac{dR}{dp} \cdot \frac{dp}{dt}.

Let me calculate the derivative and the final rate.The derivative of the revenue function with respect to price, evaluated at p=150p = 150, is:

dRdp1,510,00030112,943\frac{dR}{dp} \approx \frac{1,510,000 \sqrt{301}}{12,943}

The rate of change of revenue with respect to time dRdt\frac{dR}{dt}, considering that dpdt=4\frac{dp}{dt} = 4, is:

dRdt6,040,00030112,943\frac{dR}{dt} \approx \frac{6,040,000 \sqrt{301}}{12,943}

Let me now compute the numerical value of this.The rate at which revenue is changing when the price is $150 and increasing by $4 per month is approximately 8096.278096.27 dollars per month.

Would you like further clarification or more details on any step? Here are five related questions to expand on this:

  1. How does the product rule apply in this scenario for differentiating the revenue function?
  2. What effect does increasing price have on the number of items sold?
  3. How would the rate of change in revenue differ if the price were increasing at a different rate?
  4. What would happen if we considered the maximum price before revenue starts to decrease?
  5. How can the second derivative help in understanding the behavior of revenue with respect to price?

Tip: Always check whether a function is increasing or decreasing by examining the sign of its first derivative over the domain of interest.

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Math Problem Analysis

Mathematical Concepts

Calculus
Rate of Change
Product Rule

Formulas

x = 70000 / sqrt(2p + 1)
R = x * p
dR/dt = dR/dp * dp/dt

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Grades 11-12 or College Calculus