Use the maximum and minimum Olympiad method to do this

Let one right-angled side be x centimeters long, and the other right-angled side be (28-x) centimeters, with an area of y:

y=1/2(28-x)x

1/2(28x-x²)

1/2[-(x-14)²+196]

1/2(x-14)²+98

When x = 14 cm, the area is maximum y = 98 square centimeters.

That is, it is folded at a right angle along the midpoint of 28 centimeters, and the area is the largest, with a maximum of 98 square centimeters.

28 cm is the sum of two right-angled sides.

When the sum of two positive integers is certain, the closer the size of the two numbers is, the larger their product will be (just remember the conclusion, it's useful).

The area of a right-angled triangle is half of the product of the two right-angled sides.

So when both right-angled sides are 14cm, the area of the right-angled triangle is the largest.

s=14×14÷2=98 cm²

If you can use squares, it's better to set equations.

In fact, if this question is said to be an Olympiad, it can only be an elementary school Olympiad.

98 because the triangle area formula is.

The base times the height divided by 2

Fold a wire at a right angle, and its length is the sum of the bottom height.

So as long as the value of the base product height is guaranteed to be the largest, this triangle is the largest.

Here's a very important concept!! Note!!

When the sum of two numbers is equal, the closer they are to each other, the greater the product.

Such as 2+2=4 1+3=4 2x2>1x3 4>3 5+5=10,6+4=10,3+7=10, 5x5>6x4>3x7 25>24>23).

The sum of two numbers is known to be 28

28 is an even number, so dividing 28 by 2 equals 14

Using the area formula, divide 14x14 by 2=98

When an isosceles triangle is the largest area is 49 2

Because the rectangle has 4 lengths, 4 widths, and 4 heights, 48 4 = 12, length + width + height = 12cm, because the closer the numbers are, the greater the product, so the length, width and height are 444

x+y+z=4, find the maximum value of xyz, it's so difficult, I remember that it seems to be x=y=z=4 3 to get the maximum value, so it should be 64 27 cubic centimeters.

Personal answer (I think it should be correct): 64 cubic decimeters.

The process is as follows:

First, we cut 108 square decimeters into 108 small squares of 1 cubic decimeter.

Second, if we divide 108 cubic decimeters into 6 equal faces, then each is 108 6 = 18 (square decimeters).

Obviously, the 6 sides of the cube are squares, and we combine 18 small squares of 1 cubic decimeter into a square, and the combination of its maximum area is 4*4=16 (small squares), that is, the maximum area is 16 square decimeters, and the side length is 4.

Therefore, the maximum volume of the cube is 4*4*4=64 (cubic decimeters).

ps: The landlord gives a reward of 0 points? If so, I think, in the future, there will be no one to solve your problems for you!

Solution: Let the cube edge be x decimeter long.

6x²=108÷2

x²=9x=33³=27

Answer: 27 cubic decimeters.

Primary school Olympiad math questions, there must be a process. Compare size: a= b=

b a, because the number of digits is the same as 90876543, but the number is different, and the number of digits with a decimal place is actually doubling b. So, B is big.

First of all, I said that a must be less than b, because both numbers of a are smaller than the two numbers of b, and they are greater than 1.

I think you must have made a mistake, and you should be trying to ask: a=

Then a=b=so a-b=>0

So a>b.

Of course, there is a problem with your question, but the algorithm is probably like this.

The numerator and denominator of 333/1666 are divided by 111 at the same time: 333/1666 = 3 (15 and 1/111).

The numerator and denominator of 33/166 are divided by 11 at the same time: 33/166 = 3 (15 and 1/11).

Because 1 of 15 and 111 is less than 1 of 15 and 11, so 3 (1 of 111 of 15 and 11) is greater than 3 (1 of 15 and 11), so 333/1666 is greater than 33/166, that is, A is a big beat and B.

Because 5 squares times the 3rd power of the large state of the closed grip p is equal to the square of 3 times q, so if you want the value of c to be the smallest, the roll width p should be equal to 3, and q should be equal to 5 Suddenly I feel that I didn't say it, and I hope it can help.

(2+3) (2+6) (2+9) 4=3520 kinds 9 pages with 9 numbers, 10 99 with (99-10+1) 2=180 numbers, then there are 189 numbers within 99 pages, and if the book has x pages, then the equation (x-100+1) 3+180+9=492;x = 200 pages.

1 49 shared 5 times 5

50 59 shared 10 times 5

60 100 shared 4 times 5

Then the page numbers of the whole book are shared with (5+10+4) 2=38 3 3=18 kinds.