Question

The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of circle.​

Answer 1

Answer:

Step-by-step explanation:

K=2#r+4a (sum of perimeters)(r=radius of circle , a=side of square , # =pie)

take z=#r2+a2 (sum of products)

z=#r2+(K2+4#2r2-4#rK)/16

dz/dr=2#r+(0+8#2r2-4#K)/16=0 ( 1st derivative =0 )

32#r+8#2r-4K=0

32r+8#r-4K=0

32r+8#r-8#r-16a=0

32r=16a

2r=a ( side of square is double the radius )

find 2nd derivative by putting value of r . it will be > 0 which means it is least

hope this will be helpful