# Question Video: Finding the Maximum Area of a Rectangle given the Sum of Three of Its Sides Mathematics • Higher Education

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.

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### 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.

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

We said that if we want to maximize 𝐴, we need to find its critical points. And we’ll 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 𝐴 with respect to 𝐿. We can apply the power rule of differentiation. And it gives d𝐴 by d𝐿 is equal to 177 minus two multiplied by two 𝐿. That’s 177 minus four 𝐿. Next, we set this first derivative equal to zero and solve for 𝐿. We add four 𝐿 to each side and then divide by four, giving 𝐿 equals 177 over four or 44.25 as a decimal.

We know then that our function 𝐴 has a critical point at this value of 𝐿. 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 𝐿 into our expression for 𝐴, 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 𝐴. 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 d𝐴 by d𝐿 was 177 minus four 𝐿. So differentiating again with respect to 𝐿 gives d two 𝐴 by d𝐿 squared is equal to negative four. Now, we don’t actually need to evaluate the second derivative when 𝐿 is equal to 44.25 because the second derivative is in fact a constant. It’s the same for all values of 𝐿. 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.