<p>Some loser builds a fence around three sides of his garden. His house acts as the fourth side of the fence. What's the largest possible area of the garden if she uses 400 feet of fencing.</p>
<p>I thought about it and I got 17, 777.7777... but I don't think that is right. I got this by taking 400 and dividing into 3 because I reasoned that the two opposite sides had to be equal.</p>
<p>Then I got 20,000 by reasoning that the largest side would be opposite the house so I tried that as 200 and the other sides as 100 and got 20,000, a nice round number, but is that right?</p>
<p>You are wrong. First off, PBailey has already posted a better configuration in the first post, which also happens to be the correct answer. And michael has given you the formula to solve the problem with.</p>
<p>The maximum of the equation (400-2y)y = f(y) occurs when the first derivative is equal to 0. So f'(y) = 400 - 4y dy = 0; and therefore y = 100.</p>
<p>Y is the smaller side, so The dimensions of the rectangle would be 100 x 200.</p>
<p>khoitrinh, does that mean the derivative has to be 0 to find the maximum value for any equation like that? Is the derivative the X value? Im confused =(</p>
<p>The derivative is basically an equation representing the slope of the function at any given point. When the derivative is 0, the slope is 0 and therefore there must be a maximum or minimum.</p>
<p>But all that is calculus, which I guess most who are taking the SAT would not know. You could always graph it and calculate the maximum with your calculator as proposed by michael32, but I prefer to do it by hand because I find it to be faster.</p>
<p>It only depends on how long the loser's house is if the house is under 200 feet long. Now that I think about it, that is a fairly long house, so the problem is a little unreasonable. If the guy's house is under 200 feet, then the dimension with the largest area is a rectangle with one said that is as long as the house and the other side as long as possible.</p>