Details of materials and requiremnt of quantity in navaratnalu scheme

 

NAVARATNALU-PEDALANDARIKI ILLU - OPTION FORM

 

PEDALANDARIKI ILLU SCHEME -- HOUSES DESIGN

 

CREDITS : HOUSING DEPT, ANDHRAPRADESH GOVERNMENT

MASTER TABLE OF AGGREGATE GRADATION TABLES FOR FLEXIBLE PAVEMENTS, RIGID PAVEMENTS AND CEMENT CONCRETES

 

https://drive.google.com/file/d/11xHjHcvLF9Jqm5eOSj1kR1TjKmT31-Ou/view?usp=sharing

https://drive.google.com/file/d/11xHjHcvLF9Jqm5eOSj1kR1TjKmT31-Ou/view?usp=sharing

CREDITS:

V. VENKATA NARAYANA (EE)

N. SHASHIDAR (A.E)

HARUN KRISHNA (AEE)

VVNHIGHWAYS.BLOGSPOT.IN

Find the maximum area of the rectangular feild which is surrounded by a 100 m rope?

 Perimeter of the rectangle = 2 (L+B)

as per question 2 (L+B) = 100

                                L+B = 50

Area of the rectangle = L X B

we need to find the area of the rectangle so the maximum values of L and B must be 25, 25+25 =50 

so The maximum value is 25 X 25 = 625

                                    (Or)

L + B = 50 

B = 50 - L

Area = A = L X B

A = L X (50 - L)

    = 50 L - L^2

First derivative W.R.T  'L' gives 

dA/dL = 50 - 2L 

the value will be max when dA/dL will be minimum (or) zero

0 = 50-2L

L= 25

then B = 50 -L => B = 25