help please

1
the area of rectangle A=X (24-2X) cm^2
to find the largest possible area of rectangle, use the derivative
dA/dX
=d/dX[X (24-2X) ]
=d/dX[24X-2X^2 ]
=24-4X
let dA/dX=0
24-4X=0
X=6
when X<6 , dA/dX<0
when X>6 , dA/dX>0
so when X=6, the area of rectangle is the largest
Area
=6*(24-6*2)
=6*12
=72 cm^2

2006-11-15 13:48:44 補充：
it should bewhen X 0when X 6 , dA/dX

2006-11-15 13:49:32 補充：
when X less than 6 dA/dX greater than 0when X greater than 6 , dA/dX less than 0

以下是用 Completing Square 的方法計出答案:

Let A be the area of rectangle

A = X*(24-2X)
= 24X - 2X^2
= -2(X^2 - 12X)
= -2(X^2 - 12X + 6^2 - 6^2)
= -2[(X-6)^2 - 36]
= -2(X-6)^2 + 72

Largest possible area of rectangle is 72 when X=6.