with rectangle of a given perimeter finding the one with a maximum area and with rectangle of given area finding the one with least perimeter
Answers
Given : with rectangle of a given perimeter finding the one with a maximum area and with rectangle of given area finding the one with least perimeter
To find : prove
Solution:
Let say perimeter = P
and length = L
then Width = (P - 2L)/2
Area of rectangle = Length * width
A = L * (P - 2L)/2
=>A = LP/2 - L²
dA/dL = P/2 - 2L
dA/dL = 0
=> P/2 - 2L = 0
=> 2L = P/2
=> L = P/4
d²A/dL² = - 2 < 0
hence maximum area
L = P/4 , width = (P - 2L)/2 = P/4
Hence Square gives maximum area for a given rectangle perimeter
Given Area = A
Let say length = L
width = A/L
Perimeter P = 2(L + A/L)
dP/dL = 2 ( 1 - A/L²)
dP/dL = 0
=> 2 ( 1 - A/L²) = 0
=> L² = A
=> L = √A
width = A/L = A /√A = √A
d²P/dL² = 2 ( 0+ 2A/L³) > 0
hence minimum perimeter
Length = width = √A
Square is the rectangle with minimum perimeter for given area
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