Question

A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilibrium triangle. Find the length of each piece so that the sum of the areas of two be minimum.

Answer 1

Let s be the side of square and aa be side of equilateral triangle.

Given,

4s+3a=364s+3a=36

a=36−4s3a=36−4s3 -- ( 1 )

Total area A=s2+3√4a2A=s2+34a2

Substituting for aa from (1) above,

A=s2+3√4(36−4s3)2A=s2+34(36−4s3)2

A=s2+43√9(9−s)2A=s2+439(9−s)2

dAds=2s+43√9(2s−18)dAds=2s+439(2s−18)

d2Ads2=2+83√9>0d2Ads2=2+839>0

Hence based on second derivative test, a minimum happens for A wherever dAds=0

So,

dAds=2s+43√9(2s−18)=0dAds=2s+439(2s−18)=0

s=83√2+83√9=363√9+43√s=832+839=3639+43

Substituting in ( 1 ) and simplifying, we get

a=1089+43√a=1089+43

So size of two pieces are 4s4s and 3a3a with ss and aa with values as above.