Find Price to Maximize Revenue for 1000 Widgets

[Tekee]

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.

 

[HallsofIvy]

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.

 

[tacman]

To start, let's define our variables:

x = Price
y = Quantity Sold

We know that the company has 1000 widgets and will be able to sell all of them if the price is one dollar. This means that if the price is $1, the quantity sold will be 1000. This can be represented as the point (1,1000) on our graph.

We also know that for every 10-cent increase in price, the quantity sold decreases by 1. This can be represented as the slope of our line, which is -1/10 or -0.1.

Using the slope-intercept form of a line, y = mx + b, we can write the equation for the line representing the relationship between price and quantity sold:

y = -0.1x + b

To find the value of b, we can plug in the point (1,1000) into the equation:

1000 = -0.1(1) + b
1000 = -0.1 + b
b = 1000.1

So our final equation is y = -0.1x + 1000.1

Now, we can set up our revenue equation, which is price times quantity sold:

R = xy
R = x(-0.1x + 1000.1)
R = -0.1x^2 + 1000.1x

To find the maximum revenue, we need to take the derivative of this equation and set it equal to 0:

dR/dx = -0.2x + 1000.1 = 0
-0.2x = -1000.1
x = 5000.5

So the price that will maximize revenue is $5000.5. This may seem like a high price, but keep in mind that the company is only selling one less widget for every 10-cent increase in price. So at $5000.5, the company will sell 995 widgets, which is still a high quantity.

Therefore, the company should price their widgets at $5000.5 to maximize their revenue.