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