Apiece of wire 10 m long is cut into two pieces. one piece is bent into a square and the other is bent into an equilateral triangle. how should the wire be cut so that the total area enclosed is (a) a maximum? (b) a minimum?

a) in order to be maximum the length of the square wire should be 10m and no cut for making the triangle

b) in order to be minimum the length of the square wire should be 3.84 m and the triangle wire should be 6.16 m

Explanation:

for a square with a side length of a and a equilateral triangle of length b

area square = a²

area triangle = base* height/2

for an equilateral triangle height = base * sin 60 = (√2 / 2 )* base

therefore

area triangle = base* height/2 =  base² (√2 /4) =(√2 /4) b²

the total length of the wire is = square length + triangle length = 4*a + 3*b

therefore

A=a²+(√2 /4) b²

4*a + 3*b=L→ b = (L-4*a)/3

A=  a²+(√2 /4) (L-4*a)²/ 9

the maximum and minimum amount can be found taking the derivative of the area respect with a:

dA/da= 2*a + (√2 /36) 2*(L-4*a)*(-4) = 0

a -(√2 /9) (L-4*a) = 0

a - √2 /9 * L + √2 /9*4*a =0

(√2 /9*4+1)*a = √2 /9 * L

a= √2 /9 * L / (√2 /9*4+1) = √2 /9 * L / (√2 /9*4+1) = √2 /9 * 10 m / (√2 /9*4+1)

a= 0.96 m

therefore

b = (L-4*a)/3 = (10 m - 4*0.96m)/3= 2.053 m

A=a²+(√2 /4) b² = (0.96 m)²+ (√2 /4) (2.053m)² = 2.411 m²

in the extreme cases

a= 10 m/4=2.5 m and b=0

thus A= (2.5 m)² = 6.25 m²

b=10 m/3= 3.33 m and a=0

thus A= (√2 /4) (3.33 m)² = 3.92 m²

therefore minimum area A=3.92 m² with Length 1=4*0.96 m=3.84 m , Length 2 =2.053 m*3 = 6.16 m

the maximum area is A=6.25 m² with Length 1=10 m and Length 2=0 m