Question

If a rectangular area is required to have a perimeter of `"100 m"`, what dimensions maximize the area?

Answer 1

To do this, you have to assume a perimeter of `"100 m"`, and utilize that information to generate the largest possible area.

This can algebraically be written as:

`bb(A = lxxh)`
(area as related to length and height)

`bb(P = 2xxl + 2xxh = 100)`
(perimeter as related to length and height)

When you solve for `l` in the perimeter equation, you should get:

`2l = 100 - 2h`

`color(green)(l = 50 - h)`

If you think about the factors that multiply to give you a perimeter of `100`, pick some easy ones, and you can have:

`2xx5 + 2xx45 = 100`
`2xx10 + 2xx40 = 100`
`2xx15 + 2xx35 = 100`
`2xx20 + 2xx30 = 100`
`2xx25 + 2xx25 = 100`

Notice how if `h = 45`, then `l = 50 - h = 5`, and so on. Just pick multiple values of the height `h`, and use the corresponding value of the length `l` to look at dimension combinations.

And if you then use these lengths and heights to calculate the area you'd get:

`5xx45 = 225`
`10xx40 = 400`
`15xx35 = 525`
`20xx30 = 600`
`25xx25 = color(blue)(625)`

If you go any further, you would see that the length/height combinations have been exhausted and you would only have other symmetrical combinations (e.g. `5xx45` vs. `45xx5`).

This means the largest rectangular field in area has dimensions of `color(blue)("25 m")` `xx` `color(blue)("25 m")`.