y'=3x^2- 12x+9

Let y'=0. Then 3(x- 1)(x-3)=0.

Then the stagnation point is: x= 1, 3.

When y'>0 function is simply subtracted, y'

So when x

y″= 6x- 12。

When x= 1, y'' =-6

When x=3, y'' = 6>0, so the minimum value is: y= 1.

Y''=6x- 12. When x>2, y''>0, it is a concave interval.

Y''=6x- 12. When x

Inflexion: y''=0, so the inflexion is x=2.

That is, the circumference is 20 m. To get the maximum, because against the wall, make the long side against the wall.

If the length of the rectangle is x, then the width is: (20-x)/2.

The area is:

s= 1/2x*(20-x)≤ 1/2[(x+20-x)/2]^2=50

X=20-x at this time.

X= 10。

Bricks can build a wall 20 meters long, that is to say, the rectangle surrounded by these bricks (excluding the wall) is 20 meters long.

Total * * * can only build 20 meters. Where do you get the bricks for the remaining 20 meters of your 40 meters?