### 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.

Letβs 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 π΄ equals πΏπ. But we also have a constraint. The total length of fencing which will be twice the length plus the width β remember, the wall is making up one side of this field β 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 π΄. But before we can do that, we need to express π΄ in terms of one variable only, either the length or the width of the field. We can rearrange our constraint to give π equals 177 minus two πΏ. And then substituting this expression for π into our formula for π΄ gives π΄ equals πΏ multiplied by 177 minus two πΏ. Distributing the parentheses gives 177πΏ minus two πΏ squared. And now, we have an expression for π΄ in terms of πΏ only.

We said that if we want to maximize π΄, we need to find its critical points. And weβll 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 π΄ with respect to πΏ. We can apply the power rule of differentiation. And it gives dπ΄ by dπΏ is equal to 177 minus two multiplied by two πΏ. Thatβs 177 minus four πΏ. Next, we set this first derivative equal to zero and solve for πΏ. We add four πΏ to each side and then divide by four, giving πΏ equals 177 over four or 44.25 as a decimal.

We know then that our function π΄ has a critical point at this value of πΏ. 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 πΏ into our expression for π΄, 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 π΄. 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 dπ΄ by dπΏ was 177 minus four πΏ. So differentiating again with respect to πΏ gives d two π΄ by dπΏ squared is equal to negative four. Now, we donβt actually need to evaluate the second derivative when πΏ is equal to 44.25 because the second derivative is in fact a constant. Itβs the same for all values of πΏ. 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.