Question 881304
Please help me solve this question:
You are given two pieces of wire, each 20 cm long. One piece is to be bent to form the biggest possible rectangle and the other piece to form a golden rectangle.Determine the length and width of each rectangle.Use two methods in each case.

Note:Golden rectangle have length y cm and width x cm,where y > x,is such that  y/x = (x+y)/y = golden ratio that is 1.618
let length is Y and width  is X 
as per given first condition  
perimeter is 20 for both case 
p =2(l+b)
20= 2(Y +X)
10= Y +x  if we have pair as 1,9 area is  9    2,8 area 16   3,7  area   21   4,6 area is 24  5,5 area is 25 ( but it will be square ) therefor 
maximum area will be  24  cm^2 
so we can say biggest possible rectangle is 6,4
for golden ratio    Y>x and Y +X/Y =1.6
therefor for second  pair will be (1,9) (2,8( (3,7)(4,6) only
by which we are getting golden  ratio  Y+X/Y= 10/6 =1.66
so golden ratio shows the side should be 6 and 4 only which shows golden ratio as  length =6 cm and width =4 cm 
Answer 
biggest rectangle will be  l=6 cm and width =4 cm 
golden rectangle will  be also  6 and 4 only