Find Price to Maximize Revenue for 1000 Widgets

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SUMMARY

The discussion focuses on maximizing revenue for a company selling 1000 widgets, starting at a price of one dollar. For every 10-cent increase in price, the company sells one less widget, leading to the equation for quantity sold as 1000 - 10(x - 1). The revenue function is defined as R(x) = x(1010 - 10x), which simplifies to R(x) = 1010x - 10x². To find the maximum revenue, participants suggest using differentiation or completing the square to analyze the revenue function.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Knowledge of differentiation techniques in calculus
  • Familiarity with revenue and cost concepts in economics
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the process of finding maxima and minima using derivatives
  • Learn about completing the square for quadratic equations
  • Explore the relationship between price elasticity and revenue
  • Investigate real-world applications of revenue maximization in business
USEFUL FOR

Economists, business analysts, and students studying calculus or revenue management will benefit from this discussion, particularly those interested in optimizing pricing strategies for products.

Tekee
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A company has 1000 widgets and will bbe able to sell all of them if the price is one dollar. The company will sell one less widget for each 10-cent increase in the price it charges. What price will maximize revenues, where revenue is the selling price times the quantity sold?

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I know that I have to make a system of equations, and then multiply the equations I get for price and quantity together, all under one variable. I take the derivative of that equation and find the maximum value...

However, I do not know how to start my equations which is, unfortunately, the hard part.
 
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Well, you are told " company has 1000 widgets and will bbe able to sell all of them if the price is one dollar. The company will sell one less widget for each 10-cent increase in the price it charges. "
Okay, the number of widgets they can sell is 1000 minus the "drop off" If the price is x dollars then the increase in price (over one dollar) is x- 1. Since there are 10 10-cent increases in each dollar, that is 10(x-1) 10-cent increases in price so the number sold at price x is 1000-(10)(x-1).

10(x-1)= 10x- 10 so you can rewrite that as 1000- 10x+10= 1010- 10x. Of course, if the price is x dollars each, then the revenue is x times the number sold or just
x(1010- 10x)= 1010x- 10x^2. That's the revenue you want to maximize. Yes, you can do that by differentiating but I would consider "completing the square" to be more fundamental.
 

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