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 = 30000 / sqrt(4p + 1). Due to inflation and increasing health benefit costs, the company has been increasing the price by $5 per month. Find the rate at which revenue is changing when the company is selling widgets at $180 each.

Solution

Let's break down and solve each of the questions shown in the image.

Question 1:

We are given that the number of items sold xx depends on the price pp according to the equation: x=30,0004p+1x = \frac{30,000}{\sqrt{4p + 1}} The company is increasing the price by $5 per month, and we are asked to find the rate at which revenue is changing when the price is $180.

Steps:

  1. Revenue RR is the product of the number of items sold xx and the price pp, i.e., R=pxR = p \cdot x.
  2. Differentiate R=px(p)R = p \cdot x(p) with respect to time tt, using the product rule and chain rule.
  3. You're given that dpdt=5\frac{dp}{dt} = 5 (price increases by $5 per month), and p=180p = 180.
  4. Plug the appropriate values into the derived formula to find dRdt\frac{dR}{dt}.

Question 2:

The revenue RR is given as a function of units sold xx: R=1400x3x2R = 1400x - 3x^2 We know that sales are increasing at the rate of 40 units per day, i.e., dxdt=40\frac{dx}{dt} = 40, and we are asked to find how rapidly the revenue is increasing when 150 units have been sold.

Steps:

  1. Differentiate R=1400x3x2R = 1400x - 3x^2 with respect to time tt, using the chain rule.
  2. Plug in dxdt=40\frac{dx}{dt} = 40 and x=150x = 150 to find dRdt\frac{dR}{dt}.

Question 3:

This is a related rates problem involving two ships. At noon, ship A is 30 nautical miles west of ship B. Ship A sails west at 16 knots and ship B sails north at 16 knots. We are asked to find how fast the distance between the ships is changing at 4 PM.

Steps:

  1. Let the distance between the two ships be DD, and let the positions of the ships be modeled by the Pythagorean theorem: D2=A2+B2D^2 = A^2 + B^2, where AA is the distance traveled by ship A and BB is the distance traveled by ship B.
  2. Differentiate D2=A2+B2D^2 = A^2 + B^2 with respect to time tt.
  3. You know the speeds of both ships and their positions at 4 PM.
  4. Solve for dDdt\frac{dD}{dt}, which is the rate at which the distance between the ships is changing.

Would you like me to go into detail on any specific question? Here are 5 relative questions that can deepen your understanding:

  1. What are the rules for applying the product rule in calculus?
  2. How do we apply the chain rule when differentiating implicitly?
  3. How does related rates differ from other types of rate problems in calculus?
  4. Can you derive the revenue formula from other sales data?
  5. How can we interpret the rate of change of the distance between two moving objects?

Tip: When dealing with rates of change, always start by identifying which variables depend on time and use implicit differentiation carefully.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Differentiation
Revenue Maximization
Implicit Differentiation

Formulas

x = 30000 / sqrt(4p + 1)
R = p * x
Differentiation rules (product rule, chain rule)

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Undergraduate Calculus