### Video Transcript

A farmer wants to create a rectangular field on his land using an existing wall to bound one side. Determine, to the nearest thousandth, the maximum area he can obtain if he has 177 metres of fence to surround the other three sides.

Letโs just consider the setup here. This farmer has a wall on his land and he has a fixed length of fence, 177 metres, with which to enclose a rectangular field against this wall. The farmer wants to choose the length and width of this rectangle so that he encloses the maximum possible area. The area of a rectangle is given by its length multiplied by its width. So we want to maximize the function ๐ด equals ๐ฟ๐. But we also have a constraint. The total length of fencing which will be twice the length plus the width โ remember, the wall is making up one side of this field โ must be equal to 177. We, therefore, have an optimization problem. We have a function we wish to maximize, subject to a given constraint.

Now, in order to maximize the area of this field, we need to find the critical points of our function ๐ด. But before we can do that, we need to express ๐ด in terms of one variable only, either the length or the width of the field. We can rearrange our constraint to give ๐ equals 177 minus two ๐ฟ. And then substituting this expression for ๐ into our formula for ๐ด gives ๐ด equals ๐ฟ multiplied by 177 minus two ๐ฟ. Distributing the parentheses gives 177๐ฟ minus two ๐ฟ squared. And now, we have an expression for ๐ด in terms of ๐ฟ only.

We said that if we want to maximize ๐ด, we need to find its critical points. And weโll recall, first of all, that the critical points of a function occur, where its first derivative is equal to zero or is undefined. So we need to find the derivative of ๐ด with respect to ๐ฟ. We can apply the power rule of differentiation. And it gives d๐ด by d๐ฟ is equal to 177 minus two multiplied by two ๐ฟ. Thatโs 177 minus four ๐ฟ. Next, we set this first derivative equal to zero and solve for ๐ฟ. We add four ๐ฟ to each side and then divide by four, giving ๐ฟ equals 177 over four or 44.25 as a decimal.

We know then that our function ๐ด has a critical point at this value of ๐ฟ. But there are two more things that we need to do. Firstly, we need to evaluate the area of the field by substituting this value of ๐ฟ into our expression for ๐ด, giving 44.25 multiplied by 177 minus two times 44.25 which gives 3916.125 exactly. So we know that this is a critical point of the function ๐ด. But we must confirm that it is indeed a maximum. And in order to do this, we need to apply the second derivative test.

Recall that our first derivative d๐ด by d๐ฟ was 177 minus four ๐ฟ. So differentiating again with respect to ๐ฟ gives d two ๐ด by d๐ฟ squared is equal to negative four. Now, we donโt actually need to evaluate the second derivative when ๐ฟ is equal to 44.25 because the second derivative is in fact a constant. Itโs the same for all values of ๐ฟ. But the important thing to note is that this second derivative negative four is less than zero. The second derivative test tells us that if the second derivative of a function is negative at a critical point, then that critical point is a local maximum.

So we confirmed that this area is indeed the maximum area the farmer can obtain with 177 metres of fence. Our answer to the problem then the maximum area that the farmer can enclose with this fixed amount of fencing is 3916.125 metres squared. And this is indeed to the nearest thousandth as requested in the question.