functions question (last one)

A wire of 80cm long is cut into two pieces.Each piece of wire is bent to form a square.

a)Suppose the length of a side of one of the squares is x cm,express the total area of the two squares in terms of x.
x^2+[(80-4x)/4]^2
= x^2+(20-x)^2

b)Find the lengths of a side of the two squares so that the total area of the two squares is a minimuim.
x^2+(20-x)^2
=x^2+400-40x+x^2
=2x^2-40x+400
=2(x^2-20x+200)
By completing the square,
=2[(x-10)^2-100+200]
=2(x-10)^2+200
So, when x=10, it attains the minimum.
Thus, the length of both squares need to be 10 cm so that the total area of the two squares is a minimuim.

2007-10-05 00:37:41 補充：
part a, you may simplify x^2+(20-x)^2 into 2(x-10)^2+200我d steps清楚d~~記得揀我為最佳答案呀~~ =)

a/
The total area of the two squares
= (x * x ) + [(80 - 4x) /4]^2
= x^2 + (20 - x)^2
= (2 x^2 - 40x + 400) cm^2

b/
By method of completing square,

(2 x^2 - 40x + 400)
= 2(x^2 -20x +200)
= 2(x-10)^2 +200

The total area of the two squares is a minimuim,
when (x-10)^2 = 0,
i.e. x=10


so the lengths of a side = 10

a)
Total Area
= x^2 + [ (80-4x)/4 ] ^2
= x^2 + (20-x)^2
= x^2 + (400 - 40x + x^2)
= 2x^2 - 40x + 400
= 2 (x^2 - 20x + 100) + 200
= 2 (x - 10)^2 + 200

b)
minimum when x=10
lengths of side of the two squares are 10 and 10