Question

with rectangle of given perimeter finding the one with a maximum area and with rectangle of given area finding the one with least perimeter ​

Answer 1

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 given 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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