Hurry!! A rectangular field is enclosed with a rope 100m long

Let the length and width be a and b respectively, then a+b=100 2=50;

The first question, a*b=400, there is a*(50-a)=400=>a=10 or 40, at this time, b=40 or 10That is, when the length is 40m, the rectangular field area is 400m2

The second question, a*b=800, there is a*(50-a)=800=>a 2-50a+800=0, discriminant formula = 50 2-4*800=2500-3200<0, so there is no solution, that is, it is impossible to enclose a rectangular field with an area of 800m2.

The third question is that a*b is less than or equal to (a+b) 2 4=625, and the equal sign is held if and only if a=b=25, so the rectangular field with the largest enclosed area is 625m 2

x+y=50

x*y=400

Solve the equation to get x=40, y=10

x+y=50

x*y=800

Solve the equation: Substituting y=50-x into the lower equation yields (x-25) 2+175=0, and there is no solution.

When the circumference is constant, the area is largest when it is square. The side length is 25, so the maximum area is 625.

When enclosed in a square, the area is maximum.

At this point, the length of the rectangle = width = 100 4 = 25m

Have fun.

It is 25 meters long and 25 meters wide, with an area of 625 square meters. Let one side of the rectangle be x meters, then the other side is 100 2-x = 50-x.

s=x(50-x)

x^2+50x

x-25)^2+625

When x=25, s max=625.

Basis for solving the equation.

1. Shift the term and change the sign: move some terms in the equation from one side of the equation to the other with the previous symbols, and add and subtract, subtract and add, multiply and divide, and divide by multiplication.

2. The basic properties of the equation.

Property 1: The same number or the same algebraic formula is added (or subtracted) to both sides of the equation at the same time.

The result is still the equation. It is expressed by letters: if a=b, c is a number or an algebraic formula.

1)a+c=b+c

2)a-c=b-c

Property 2: Multiply or divide both sides of the equation by the same non-0 number, and the result is still the equation.

It is expressed in letters as: if a=b, c is a number or an algebraic formula (not 0). Then:

a c = b c or a c = b c

Property 3: If a=b, then b=a (symmetry of the equation).

Property 4: If the god is a=b, b=c, then a=c (the transitive nature of the equation).

Let the length be x, the width will be 50-x, x<50-x> the number of cherry blossoms should be greater than 600, and the length and width are limited, and the laughter is solved by mean inequality.

It is 25 meters long and 25 meters wide, with an area of 625 square meters.

Let the long side of the enclosed rectangle be x, then the short side is (and filial piety 10-x), so s=x(10-x)=-x-5) 2 +25, and the area formula has a shed roll function image opening downward

When x = 5, the maximum area is 25 m2

So the answer is: 25m 2

When the circumference is constant, the square area in the quadrilateral is the largest.

y1=（a/4）^2=a^2/16

If the length of the rectangle is x when the area is maximum, then the width of the friend is 50-xs=x(50-x) and the good servant = 50x-x2

Its maximum is obtained at x = -50 -2 = 25, where the area is 25 * 25 = 625 square centimeters.

The length and width are 25 cm and 25 cm respectively

Length and width 120 4 = 30 (cm).

A: When both the length and width are 30 cm, the enclosed rectangle has the largest area.

It can be enclosed in a rectangle 6 meters long and 4 meters wide.

The process is as follows:

1) Let the length of the rectangle be xm, then the width is (20 2-x)m, which is derived from the title:

2）x（20÷2-x）=24

3）x（10-x）=24

4) Solve x1=4 (discarded if not in accordance with the topic), x2=6. So the length of the rectangle is 6m, and the width is 10-6=4m.

A: It can be enclosed into a rectangle with a length of 6 meters and a width of 4 meters.

Let the length of the rectangle be xm, then the width is (20 2-x)m, and x(20 2-x)=24 from the title

x（10-x）=24

The solution is x1 = 4 (discarded if it doesn't fit the topic) and x2 = 6, so the length of the rectangle is 6m and the width is 10-6 = 4m

A: It can be enclosed into a rectangle with a length of 6 meters and a width of 4 meters.

1: The side length of the square is l 4 s l 4 l 425 ll 16

l 20 so: l 20 cm

2：2πr＝l

r＝l/2π

s＝πrrs＝π＊ll/（2π＊2π）

100＝πll/4ππ

400ππ＝ll

400π＝ll

l＝20√π

So: l 20

3: l 8 s round ll 4 16 s square: 4l 12 s round ll 4 36 s square: 94: so the perimeter is the largest area enclosed in a circle when the value is fixed.

If the area of the square is not more than 25 square centimeters, then the rope length l should meet l 5;

2) If the area of the circle is not less than 100 square centimeters, then the length of the rope l should meet l 20

3) When l=8, the area of the circle is large, and when l=12, the area of the circle is large;

4) What conjectures can you get? The perimeter is the area that encloses the circle at a fixed value. ＿＿

Less than or equal to 5

Greater than or equal to 20 followed signs

3.Rotundity; Rotundity;

4.The area of the graph with equal side lengths is not necessarily the same; When the circumference is equal, the area of the circle is the largest.

5.Basic properties of the equation: 1. The same number is added or subtracted from both sides of the equation, and the equation is true; 2. Multiply or divide by the same number on both sides of the equation, and the equation holds.

If the length is x, the width will be 10-x

From the meaning of the question, x(10-x)=24 is solved.