You can sell 800 vacuums for $100 each. For every one dollar you increase you can sell two fewer vacuums. For example, if you charge $103 you can sell 794= 800-2(3) vacuum cleaners. If you charge 98, you could sell 804= 800 + 2(2) vacuum cleaners. How much should you charge to maximize revenue?
You are on the right track. We can find a linear equation to predict the number of vacuums that will be sold at any given price. At $100 it is 800. At $103, it is 794. Think of this data as two points on a line. The points are (100,800) and (103,794). Find the equation of the line. y=1000-2x. The Revenue generated is the number sold * the price. y is the number sold and x is the price. R=x*y=x*(1000-2x) R(x)=1000x-2x? R'(x)=1000-4x The critical number is when R'=0 == 1000-4x=0 == x=250 So, it looks like we should charge $250 per vacuum. Is this really a maximum? R''(x)=-4. The Second Derivative Test tells us that our value is a maximum. This makes sense. The revenue function is a parabola that opens downward with a vertex at (250,125000), The price is $250 and the revenue is $125000.
