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

Question

Kennedy Adams · Accepted Answer

To do this, you have to assume a perimeter of #"100 m"#, and utilize that information to generate the largest possible area. This can be expressed algebraically 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 computed the area using these heights and lengths, you would 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")#.

Eli Assouline · Answer

undefined The dimensions that maximize the area of a rectangular area with a perimeter of 100 m are when the rectangle is a square. In this case, each side would be 25 m, resulting in a maximum area of 625 square meters.