Area: 64 square feet
2007-10-30 09:32:49 · 7 answers · asked by nala 1 in Science & Mathematics ➔ Mathematics
Hello,
P = 2L + 2W
A = LW so 64 = LW or W = 64/L. Now put this into the perimeter equation and we have P = 2L + 2(64/L) = 2L + 128/L
Take the derivative and we have DP/DL = 2 + -(1)128/L^2 now to get the minimum set it equal to 0 giving us 0 = 2-128?l^2 multiple by L^2 so we have 0 = 2L^2 -128 so L^2 = 64 and L = 8 ft> and W = 8 ft.
Hope This Helps!!
2007-10-30 09:43:09 · answer #1 · answered by CipherMan 5 · 0⤊ 0⤋
The rectangle with the minimum perimeter is a square. Since that area is 64 square feet, that means that if it were square each side would be 8 feet. So the minimum perimeter is 8 feet x 4 = 32 feet.
2007-10-30 16:38:59 · answer #2 · answered by Q Girl 5 · 0⤊ 0⤋
Area=64
So the possible values for length and breadth can be
l=32 b=2
l=16 b=4
l=8 b=8
Since it is said that the perimeter is minimum,
l=8 b=8,satisfies the above condition.
perimeter=2*(8+8)=32
2007-10-30 16:39:32 · answer #3 · answered by d p 1 · 0⤊ 0⤋
Hint: the minimum perimeter rectangle is a square.
.
2007-10-30 16:36:27 · answer #4 · answered by tlbs101 7 · 1⤊ 0⤋
Length = 8 feet
Width = 8 feet.
The perimeter is 32 feet.
2007-10-30 16:35:53 · answer #5 · answered by jlao04 3 · 1⤊ 0⤋
There are only 4 possiblities, and they are directly related to factors of the area.
2007-10-30 16:38:35 · answer #6 · answered by solomon d 2 · 0⤊ 1⤋
Do you own homework
2007-10-30 16:35:13 · answer #7 · answered by Fuzzybutt 7 · 0⤊ 2⤋