Maximizing Revenue by Analyzing a ParabolaDate: 03/13/2007 at 16:41:50 From: Chryssy Subject: need help with quadratic word problem When priced at $30 each, a toy has annual sales of 4000 units. The manufacturer estimated that each $1 increase in cost will decrease sales by 100 units. Find the unit price that will maximize total revenue. I am getting very frustrated with this problem. My end answer for the number of units isn't right and I don't know what I'm doing wrong. -100x^2 + 1000x + 120000 = 0 100x^2 - 1000x - 12000 = 0 100(-x^2 + 2*5x + 2^4*3*5^2) = 0 100(-x^2 + 10x + 16 - 3 - 25) = 0 100(-x^2 + 10x + 1200) = 0 -x^2 + 10x + 1200 = 0 x^2 - 10x - 1200 = 0 I worked the quadratic equation and came up with x = 40,-30. I throw out -30 and insert 40 to determine the answer. That makes the price of the toy = $30 + $1(40)= $70.00, and the total number of units = 4,000 - 100(40) = 0. But that means selling 0 units, which can't be right. What am I doing wrong? Date: 03/13/2007 at 20:08:35 From: Doctor Douglas Subject: Re: need help with quadratic word problem Hi Chryssy. Your equation is correct, but you're making a key mistake in interpreting its meaning. I'll assume that you did something like the following to come up with your equation. The total revenue that you want to maximize is R = (30 + x) * (4000 - 100*x) where x is the potential change in dollars from the base price. The first factor is the price per unit, and the second factor is the number of units. R = 120000 + 1000x - 100x^2 Now, what should we do with this equation? The key thing to keep in mind is that the equation represents the revenue. For example, when x = 0 (no change in the price), we get R = 120,000 dollars revenue since they sell 4000 units at $30 each. Since you set your equation equal to zero and then solved it, you actually found the number of units to sell to make the revenue be 0, which is what happened with a $40 increase, right? You charge $70 but nobody buys any. The same result happens when x = -30 since that makes the price per unit be $0 and you won't make much money that way! You want to find where R is maximized, not where it becomes zero. Now, because the graph of R vs. x is a parabola that opens downward, and because you found that it crosses the x-axis at -30 and 40, you can make an educated guess that right in the middle of those two points (i.e., at x = 5), R will be maximized. You can confirm this by graphing R versus x. There are other methods for finding the maximum or minimum (which occurs at the vertex) of a parabola, but I like this idea of using the value midway between the x-intercepts. You can read about some other approaches to a similar problem here: Pumpkin Price http://mathforum.org/library/drmath/view/53269.html Write back if you have questions about this. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 03/15/2007 at 10:32:05 From: Chryssy Subject: Thank you (need help with quadratic word problem) Thank you so much, Dr. Douglas! Now it makes perfect sense. Thank you for your prompt reply as well. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/