Optimization and maximum area of a rectangular enclosure

[TSN79]

I've used differentiation to find that a rectangular enclosure made up of a 100m fence should have four sides all 25m to be as large as possible. The function I get is 50x-x^2. As I said, differentiating this function gives me the largest area possible. But how would I go about finding how long the sides must be in order to make the area as small as possible...?

 

[vsage]

the maximized area was found at the maximum value of x (which happened to fall on the local maximum) so it makes sense the minimized area would be found at a value of x at the absolute minimum on the acceptable domain. Just be sure to keep within the constraints of what the perimeter is.

 

[wais]

To find the minimum area of the rectangular enclosure, we can use the same approach of differentiation. However, instead of finding the maximum value, we will find the minimum value of the function.

Let's first set up the function for the area of the rectangular enclosure:

A = xy

Where x is the length and y is the width of the enclosure. We know that the perimeter of the enclosure is 100m, so we can set up an equation to represent this:

2x + 2y = 100

Solving for y, we get:

y = 50 - x

Substituting this into the area function, we get:

A = x(50 - x) = 50x - x^2

To find the minimum value of this function, we can use the process of differentiation. Taking the derivative of the function, we get:

A' = 50 - 2x

Setting this equal to 0 to find the critical points, we get:

50 - 2x = 0

2x = 50

x = 25

We can see that this value of x is the same as the one we found for the maximum area. This makes sense because the minimum and maximum values of a quadratic function occur at the same x-coordinate.

To find the minimum area, we can plug this value of x back into the original function:

A = 50(25) - (25)^2 = 625

Therefore, the minimum area of the rectangular enclosure is 625 square meters when the sides are both 25 meters long. This result is also supported by the fact that a square has the largest area for a given perimeter. In this case, the square has sides of 25 meters and an area of 625 square meters.

In conclusion, by using differentiation, we can find both the maximum and minimum values of a function, in this case, the area of a rectangular enclosure. This allows us to optimize the dimensions of the enclosure to either have the maximum or minimum area.