 Hi everyone, let's take a look at another example of an optimization modeling problem. Here we have a farmer who wants to fence in 60,000 square feet of land in a rectangular plot, then divide it in half with a fence parallel to one pair of sides. And we are asked to determine what the dimensions are of the plot that will require the least amount of fence. Let's just draw ourselves a rectangle to represent the farmer's plot of land. And we know we want to divide it in half. It doesn't really matter if you divide it, you know, up or down or across. I'm just going to do it vertically just to make it easier. For some reason I think that's easier. And I tried to make it in half. If you think of the steps that we want to follow, so we have here a rough diagram of what this would look like. Let's start assigning some variables to this. Perhaps we'll call this length here x, the length of the entire plot, and we'll refer to the height this way as y. And of course we know this fence that the farmer put in the middle that has a length of y as well. So that would be our diagram. Let's go ahead and figure out what it is we were given. We're told that we have an area of 60,000 square feet. And of course we know areas length times width. So that would be x times y. But we're looking to minimize, because remember we're trying to determine the least amount of fence that can be used. So that is a minimization problem. And so we want to minimize the amount of fence. We need to come up with an equation to model how much fence would be used. So I'm going to call that f for fence. This is essentially a perimeter equation that we're going to write because we'd have two lengths of length x. So that's 2x. So that accounts for the fence along the bottom and along the top. Then we have these three vertical lengths. So that would make plus 3y. Thinking again of the steps we go through to solve an optimization problem, this would be our primary equation. This f equals 2x plus 3y. Of course the problem is it's in terms of two different variables which we cannot have. So we need to go back to that other equation we wrote, our secondary equation that models the area and solve for either x or y and then substitute into the f equals 2x plus 3y equation. Suppose we just solve for y. That would give us y equals 60,000 over x and we are going to substitute that in place of the y in the primary equation. So now we have a new primary equation, f equals 2x plus 3 times 60,000 over x. And we'll go ahead and simplify that so that gives us 2x plus 180,000 over x. And that becomes the primary equation that we are going to seek to minimize. We know our next step will be to take the derivative of this new primary equation. So our derivative would be 2 minus, now if we think of this 180,000 x to the negative 1, our derivative then becomes minus 180,000 over x squared. It would have been x to the negative second so I'm just going to rewrite it as x squared. We know that in order to find a minimum of a function we set that derivative equal to 0 and we are going to go ahead and solve that. Solving that equation yields a critical number of 300. So we know that if there is going to be a minimum it will occur at this value of 300. We need to go ahead then and verify that this is indeed a minimum. We can do that with a number line analysis. Remember that you're substituting values into the derivative into f prime. We know it's at 300 that derivative equals 0. If we were to choose a number less than 300 we know we can't go below 0 because remember what our x's stand for. Our x's represent a length of fence so obviously you cannot have a negative length. We know we cannot go lower than 0. Maybe let's pick like 100 or something like that. If we substitute that in place of x in here we do end up with a negative value. You can go ahead and verify that on your calculator. If we choose something bigger than 300 and substitute into x here maybe like 500 or something we do get a positive value. Therefore we have our derivative changing from negative to positive meaning the original function is switching from decreasing to increasing therefore creating a relative minimum at 300. The only thing we have left to do then is to actually answer the question. We now know that this length of x has to be 300 feet. We need to find the width, the y. The easiest place to go is to go back to that y equals 60,000 over x substitute 300 and 4x there and you should find that y equals 200. We find then that in order to minimize the total amount of fence the farmer's plot should measure 300 feet by 200 feet.