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

##### 1 Answer

To do this, you have to assume a perimeter of

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

#2l = 100 - 2h#

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

If you think about the factors that multiply to give you a perimeter of

#2xx5 + 2xx45 = 100#

#2xx10 + 2xx40 = 100#

#2xx15 + 2xx35 = 100#

#2xx20 + 2xx30 = 100#

#2xx25 + 2xx25 = 100#

Notice how if

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.

This means the largest rectangular field in area has dimensions of