Primary 6 questions about the volume of the surface area of a cuboid

Question 1: To make the volume maximum, it is only when it is a cube, let the side length of the cube is x, then the side length of the subtracted cuboid is 24-2x, the side length is the height of the cube, then they are equal, we can know that x is 8, so the side length subtracted is 24-2x=8 Question 2: Let the cube with bricks as 1, then the height of the cuboid is 10, the width is 7, and the length is 38, and there are only three sides that are brushed, and the area of one side of each brick is 1, and the total number of blocks is 38*7*10=2660, the fast number of the left brush has 38*10=380, above:

38*7=266, and the other is 7*10=70, so there are 1844 that are not brushed

Question 1. Let the subtracted side length be x

The bottom surface of the carton is 24-2x in length and width

24-2x>0 x<12

The carton volume is v=(24-2x)(24-2x)x=4x(12-x)(12-x).

When 4x=12-x, the carton volume is maximum.

Solve the equation to get x=12 5

Question 2. There are 38 bricks 7 10 = 2660 bricks.

The top layer is all brushed with lime water, and there are 38 7=266 pieces.

Both sides are all brushed with lime water, because the top layer has been calculated, and the less layer is counted as 9 layers.

There are 37 9 + 7 9 = 396 blocks.

There are 2660-266-396=1998 blocks that have not been brushed.

Finally, I wish the landlord a happy day.

v=x(24-x)^2=x(576+x^2-48x)x^3-48x^2+576x

v'=3x^2-96x+576=0

x1=24 (discarded if not in place).

The side length of the small square minus x2=8 should be 8 cm.

The top is exposed, with 1 side of the length and 1 side of the width exposed, so 38-1) (7-1) (10-1) = 1988

Question 1: Let the side length of the cut small square be x cm.

Volume v=x(24-2x) 2

4(x^3-24x^2+144x)

v'=4(3x2-48x+144)=12(x-12)(x-4)lingv'=0 then x=12 (non-conformity) x=4

So the side length of the small square should be 4cm

Let the side length of the transformed cube be a

Rule. Reduced surface area = 4 (3a + 2a) = 120 so. a=6

Protocuboid. The length is 6 + 3 + 2 = 11

The width and height are both 6

So the volume is 6x6x11=396

Let the cube side length be x, then.

The reduced surface area is 3x*2+3x*2+2x*2+2x*2=120, i.e. 20x=120

x = 6 (cm).

The length of the original cuboid: 6 + 3 + 2 = 11 (cm).

Width: 6 (cm).

Height: 6 (cm).

Volume: 11*6*6=396 (cubic centimeters).

120 (3+2) 4=6 cm. 3+2 means that the left and right boxes are combined into one. This 6 is both the edge length of the cube that becomes and the width and height of the original cuboid.

6+2+3 11 cm. 11 is the total length of the original cuboid.

11*6*6 396 cubic centimeters.

According to the inscription, the left and right sides of this cuboid are squares, and the side length is:

120 4 (3+2) = 6 cm.

Then the original length of this cuboid is: 3 + 2 + 6 = 11 cm.

The original cuboid volume is: 11 6 6 = 396 cubic centimeters.

Volume 6 * 6 * 60 = 2160 cubic centimeters.

Surface area 2 * (6 * 6 + 6 * 60 + 6 * 60) = 1512 square centimeters.

From 96 4 = 24, the area increased on each side is 24 square centimeters.

So the square side length is 24 4 = 6 cm.

The volume and surface area are then obtained by the formula.

The increase in surface area is equivalent to the area of a box with a height of 2 meters and no top and bottom, so the perimeter = 64 2 = 32 meters, that is to say the side length = 32 4 = 8 meters.

It turned out to be a cube of the pit was put and then dug 2 meters, so the size of this rectangular pit is.

8*8*10=880 cubic meters.

5-2=3

One side: (3 2 + 3 1 + 2 1) 2 = 22 Two sides: (3 + 2 + 1) 4 = 24

Three-sided: 8 colorless: 3 2 1 = 6

The first three-sided ones are at a vertex, there are 8 in total; The two sides are on each edge, and the three sides are removed, :(5-2+4-2+3-2)*4=24

One side on each side of the box: (5-2) (3-2) * 2 + (5-2) (4-2) * 2 + (4-2) (3-2) * 2 = 22

None of the two sides are painted: 5*4*3-24-22-8=6

(1) 3 cm = m.

Volume = square meters.

2) The volume of iron beads = 96 * 2 = 192 square meters, and at least 2 beads should be placed.

A total of three water grains are needed to overflow.

This is obviously a fifth grade, which is easy to do, but now I am also in a hurry to do my homework.