how do you find the maximum area of a rectangle with a perimeter of 44 centimeters?

Question

and the same for 26 meters and 5 millimeters? thank you!

Answer 1

2L + 2W = 44
L + W = 22
W = (22 - L)

L x W = Area
L x (22-L) = Area
22L- L^2 = Area
22 - 2L = 0

L = 11 and W = 11 max area 121 square centimeters

2L + 2W = 26 meters
L + W = 13 meters
W = (13 - L)
13L - L^2 = 0
13 - 2L = 0
L= 6.5 W=6.5 and max area = 42.25 sq meters


2L + 2W = 5
L + W = 2 1/2
W = (2 1/2 - L)
L (W) = Area
L ( 2 1/2 - L ) = Area
2 1/2 L - L^2 = Area

2 1/2 - 2L = 0
L = 1 1/4 mm W = 1 1/4 mm Area 1 1/16 sq mm


Just what kind of rectangle has a perimeter of 5 mm except one in a math book?

Answer 2

Maximum area is obtained when all sides are equivalent. Practically the rectangle turns into a square though. All sides become 11 centimeters and you do 11 x 11 and get your answer to be 121 cm^2

For 26 meters the maximum area would be (26/4) x (26/4)

For 5 milimeters the maximum area would be (5/4) x (5/4)

Answer 3

the max area is when it's a square.
here is the proof
A=LxW,
L+W=44/2 => W=22-L
A=LX(22-L)
dA/dW=0 to get the max/min
22-2L=0
2L=22
L=11 then W=11
A=11x11=121

Answer 4

2l+2w=44
2w=44-2l
w=(44-2l)/2 -- (1)
lw is area
l(44-2l)/2
=22l-l^2
dA/dl =22-2l=0
l=11
d^2A/dl^2=-2 < 0
w=11 (from 1)
Area = 121 sqauare centimeters