F4.certificate Maths功課...幫手!!!快快快...

Cut a wire of length 80cm int 2 parts,and then change them to square respectively.The length of each side of the smaller is 4x cm.
(a) Respresent the total area of the two spuares A in terms of x.
這兩個正方形的面積為
x^2 + [(80 – 4x)/4]^2
= x^2 + (20-x)^2

(b) Find the minmum value of A.
f(x) = x^2 + (20-x)^2
f(x) = x^2 + 400- 40x + x^2
f(x) = 2x^2 - 40x + 400
f(x) = 2(x^2 - 20x) + 400
f(x) = 2(x^2 - 20x +100 - 100) + 400
f(x) = 2[(x +10) ^2 – 100] + 400
f(x) = 2(x +10) ^2 - 200 + 400
f(x) = 2(x +10) ^2 + 200
所以最小面積為 200cm^2 當 x = 10cm

(a)
Let 16x cm be the length of shorter part of wire
then another part is (80-16x)cm

The length of each side of the smaller is 4x cm
The length of each side of the larger is (20-4x) cm

A=(4x)^2 + (20-4x)^2
=16x^2 + 400 - 160x + 16x^2
=32x^2 -160x + 400

(b)
A=32x^2 -160x + 400
=32(x^2 - 5x) +400
=32[(x-5/2)^2 - 25/4] +400
=32(x-5/2)^2 + 200
Therefore, the minmum value of A is 200cm^2 when x = 5/2