Find the maximum area of a rectangular piece of land that can be surrounded by a fence that is 12 meters long.
The question is asking us to maximize the area of a rectangular piece of land which is surrounded by a fence that is 12 meters long. Let’s start by sketching the information given to us in the question. We have a rectangular piece of land. Let’s call the length of this rectangle 𝐿 and the width 𝑊. And the area of this rectangle is the length multiplied by the width. And remember, the question wants us to maximize the value of this area. Also, remember we’re told the length of our fence needs to be 12 meters. And of course, the length of our fence will just be equal to the perimeter of our rectangle, which we can find by just adding the lengths of all the sides. So we get two 𝐿 plus two 𝑊. And this is equal to 12 meters.
So we’ve been asked to maximize our area subject to two 𝐿 plus two 𝑊 is equal to 12. This is an optimization problem. So we’ll want to solve this by finding the critical points of our area function. However, as it stands, our area function is a function of two variables. So our first step will be to rewrite the area as a function of one variable. We can do this by rewriting the equation two 𝐿 plus two 𝑊 is equal to 12 to make either 𝑊 or 𝐿 the subject of this equation. It doesn’t matter which one we’ll pick. We’ll make 𝐿 the subject of this equation. We’ll start by dividing through by two. This gives us 𝐿 plus 𝑊 is equal to six. Finally, we’ll subtract 𝑊 from both sides of this equation. We get 𝐿 is equal to six minus 𝑊.
Now, we want to substitute 𝐿 is equal to six minus 𝑊 into the equation for our area. This gives us that the area of our rectangle 𝐴 is equal to six minus 𝑊 times 𝑊. And we can distribute 𝑊 over our parentheses to get six 𝑊 minus 𝑊 squared. Now, to maximize the area of our rectangle, we’ll want to find the critical points of this function. And this is a polynomial, so its critical points will be where its derivative is equal to zero. So let’s find the derivative of our area with respect to 𝑊. We get that this is equal to the derivative of six 𝑊 minus 𝑊 squared with respect to 𝑊. And we can do this term by term by using the power rule for differentiation. We get six minus two 𝑊.
Now, we’ll solve this equal to zero. This will give us the critical point of our area function. To solve this, we add two 𝑊 to both sides and then divide through by two. We get that 𝑊 is equal to three. So 𝑊 is equal to three is a critical point of our area function. Remember, we still need to check what type of point this is. We could use the first or second derivative test. However, in this case, our area function is a quadratic. And in this case, it’s a quadratic where the leading term has a coefficient of negative one. So we know the general shape of this function. It’s a parabola with a negative leading coefficient.
So in actual fact, our turning point is not only a local extremer; it’s a global extremer. And we know there’s only one turning point. It’s the critical point where 𝑊 was equal to three. So 𝑊 equals three is a global maximum of our area function. But remember, the question wants us to find the area of our rectangle. So we’ll substitute 𝑊 is equal to three into our expression for the area. Substituting 𝑊 equals three into our formula for the area, we get six minus three times three, which we can calculate to give us nine. And of course, since we were using meters as our units for the length, our area will have the units of meters squared.
Therefore, we’ve shown the maximum area of a rectangular piece of land that can be surrounded by a fence which is 12 meters long is nine meters squared.