Question Video: Finding the Rectangle Dimensions That Would Yield the Maximum Area given Its Perimeter Using Differentiation Mathematics • Higher Education

A wire of length 41 cm is used to make a rectangle. What dimensions give its maximum area?

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Video Transcript

A wire of length 41 centimeters is used to make a rectangle. What dimensions give its maximum area?

Now, it’s important that we don’t just attempt trial and error here. We need to use a proper optimization method to find the dimensions which will give the largest possible area for this rectangle, subject to the constraint that we only have 41 centimeters of wire.

Let’s consider then a rectangle with a length of 𝑙 centimeters and a width of 𝑤 centimeters. We need to maximize its area, which for a rectangle is its length multiplied by its width. Subject to the constraint which is that the perimeter of this rectangle must be equal to 41. The perimeter of a rectangle is twice its length plus twice its width. So, we have the constraint two 𝑙 plus two 𝑤 equals 41.

Now, in order to maximize this area, we’re going to need to use differentiation to find the critical points of this function 𝐴. But before we can do that, we need to write 𝐴 in terms of one variable only. The choice of whether we use 𝑙 or 𝑤 is entirely arbitrary. So, I’ve chosen to rearrange the linear equation to give 𝑙 equals 41 minus two 𝑤 over two. Substituting this expression for 𝑙 into our area formula gives 𝐴 equals 41 minus two 𝑤 over two multiplied by 𝑤. And distributing the parentheses, we have 41𝑤 over two minus 𝑤 squared.

To find the critical points of 𝐴, we first find its first derivative, d𝐴 by d𝑤, which, using the power rule of differentiation, is equal to 41 over two minus two 𝑤. We then set this expression equal to zero and solve the resulting equation for 𝑤. We add two 𝑤 to each side and then divide by two, giving 𝑤 is equal to 41 over four. So, we’ve found the width of the rectangle at which the area has a critical point.

We also need to find the length though, which we do by substituting this value of 𝑤 back into our expression for 𝑙, giving 41 minus two times 41 over four all over two. Which also simplifies to 41 over four. However, we’re not yet finished. We know that these values of 𝑤 and 𝑙 give a critical point for the area. But we haven’t yet confirmed that it is indeed a maximum. To check this, we need to perform the second derivative test. We find d two 𝐴 by d𝑤 squared, which is equal to negative two.

Now, this is a constant for all values of 𝑤. But more specifically it is a negative constant. And as the second derivative is less than zero, this confirms that our critical point is indeed a maximum. So, we found that the length and width which give this rectangle its maximum area, subject to the given perimeter constraint, as decimals, are both 10.25 centimeters.