Maximizing OrangesDate: 06/29/2002 at 09:27:37 From: Julie Subject: maximizing An orchard has 800 orange trees, each of which yields 120 oranges per season. For each new tree that is added to the orchard, the output per individual tree decreases by 2 oranges per season. Determine the number of trees that would maximize the orchard's output. I have no idea where to begin! Date: 06/30/2002 at 00:25:24 From: Doctor Achilles Subject: Re: maximizing Hi Julie, Thanks for writing to Dr. Math. This is a problem that requires you to use a lot of different skills, so it's really tough, including some calculus. Let's take it step by step. So you already know that you need to maximize the number of oranges. Let's call the total number of oranges "n". So we want to maximize n (make it as big as we possibly can). How do we find n? In other words, what equation can we use to figure out what n is equal to? Well, n is the total number of oranges. The way the orchard started out, there are 800 trees and each tree yields 120 oranges. With that set up, what is n? Why, it's just 800 * 120 (the "*" is a multiplication symbol). So we can find what n is with 800 trees, but the problem is trickier than that. We can go changing the number of trees. Since the number of trees is variable (it changes), let's call it "t". It gets even more complicated though, because not only does the number of trees change, so does the yield of oranges per tree. Let's call the yield per tree "y". Remember how we found n. We found it by multiplying the number of trees by the yield per tree. So we can write down this equation: n = t * y The values of n, t, and y might change, but the the equation is always going to be true. The next thing to do is figure out how y changes when we change the number of trees. To do that, let's just forget about n for a while and try to find an equation that relates y and t to each other. Here's what we know: 1) y = 120 when t = 800 2) When you add 1 to t, you have to subtract 2 from y. So: y = 118 when t = 801 And y = 116 when t = 802 etc. If we wanted to, we could graph the relationship between t and y: y | | | | | | | --------------|------------------ t | | | | | | Where could we put a point? Well we could put a point at (800, 120) and another at (801, 118) and another at (802, 116). These points will all form a line. Now, we need to find an equation for the line. The first step is to find the slope of the line. To do that, we take the rise (change in y) divided by the run (change in t). (116 - 120) -4 m = ------------- = ---- = -2 (802 - 800) 2 So the slope is -2. So our equation will look like: y = -2*t + b ...where b is the y-intercept. To find b, we use the point (800, 120) and plug it in to the equation: 120 = -2*800 + b Which simplifies to: 120 = -1600 + b or: 120 + 1600 = b or: 1720 = b So our equation is: y = -2*t + 1720 We now have these two equations: n = t * y y = -2*t + 1720 We can use substitution on y. In other words, we can substitute (-2t + 1720) for y in the first equation. That will give us: n = t(-2t + 1720) If we multiply the t over the parentheses, we get: n = -2t^2 + 1720t (The little ^2 means "squared", the "^" is for an exponent.) So now we need to find the maximum value of the expression: -2t^2 + 1720t There are a few different ways to do this. Here are two I can think of: 1) The easy way is to use a graphing calculator to graph the equation and find the maximum. 2) Since the equation is quadratic, it will only have one point where the slope is equal to zero. This will be the maximum of the function (if it has a maximum) or the minimum of the function (if it has a minimum). You can see the way this works by looking at a graph of the equation on a graphing calculator. The point where the slope goes from being positive (sloping up) to being negative (sloping down) is the maximum, at that point the slope is zero. Remember that the way to find the slope of a function is to take the derivative. If you want a quick review of derivatives, check out this page in the Dr. Math archive: Derivatives http://www.mathforum.com/library/drmath/view/53398.html The derivative of the function -2t^2 + 1720t is -4t + 1720 We want to find a point where the slope is zero (and therefore we want a point where the derivative is zero). So we want to solve this equation: 0 = -4t + 1720 or 4t = 1720 or t = 430 So the magic point is where t = 430 -- in other words, where the number of trees is equal to 430. Just to double check, you should go back and figure out how many oranges there will be when there are 430 trees. To do that, first figure out how many oranges each tree will be making, then figure out what the total number of oranges will be. Next, check how many oranges will be produced by 431 trees using the same procedure. Then check how many will be produced by 429 trees. If 430 is the right answer, then both of these numbers should be smaller. [NOTE: I did NOT check my work, so I could be wrong.] Hope this helps. If you have other questions about this or you're still stuck on any point, please write back. - Doctor Achilles, The Math Forum http://mathforum.org/dr.math/ |
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