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, itβs important that we donβ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.

Letβs consider then a rectangle with a length of π centimeters and a width of π€ 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 π plus two π€ equals 41.

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

To find the critical points of π΄, we first find its first derivative, dπ΄ by dπ€, which, using the power rule of differentiation, is equal to 41 over two minus two π€. We then set this expression equal to zero and solve the resulting equation for π€. We add two π€ to each side and then divide by two, giving π€ is equal to 41 over four. So, weβ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 π€ back into our expression for π, giving 41 minus two times 41 over four all over two. Which also simplifies to 41 over four. However, weβre not yet finished. We know that these values of π€ and π give a critical point for the area. But we havenβt yet confirmed that it is indeed a maximum. To check this, we need to perform the second derivative test. We find d two π΄ by dπ€ squared, which is equal to negative two.

Now, this is a constant for all values of π€. 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.