The length of the midline CD on the waist AB of the isosceles triangle ABC is 2, then the maximum ci

Linear programming. Let ab=ac=2x, bc=y, known cd=2, and the circumference of the triangle abc z=4x+y, which can be seen from the trilateral relationship of the triangle.

3x>2

x<2x+y>2

x-y<2

4x>y

x>0y>0 makes a feasible domain in a Cartesian coordinate system, and for the objective function z=4x+y, z is taken to a maximum value of 16 at (2,8).

The formula for the length of the middle line, let the waist be 2x and the bottom be y, then 2x 2+y 2=8 circumference = 4x+y

Here you can use the Cauchy inequality directly.

2x 2+y 2)(8+1) (4x+y) 2 so 4x+y 6 2

So the maximum value is 6 2

Now verify that it matches the topic.

x=2y fits the topic (use your brain).

In the case of an equilateral triangle, the circumference is the shortest.

Let's do the math for the specific answer, first assume that it is an equilateral triangle, and then prove that it is the maximum.

The proof process is not written, and the trigonometric function is used to prove it.

The length of the high cd on the AB side is 2

As shown in the figure below, in a right triangle with an angle of 30°, the ratio of the length of the three sides is 1:2: 3cd = 1 2 ac = 1 2 ab = 2<>

Linear programming. Let ab=ac=2x, bc=y, known cd=2, and the circumference of the triangle abc z=4x+y, which can be seen from the trilateral relationship of the triangle.

3x>2

x<2x+y>2

x-y<2

4x>y

x>0y>0 makes a feasible domain in a Cartesian coordinate system, and for the objective function z=4x+y, z is taken to a maximum value of 16 at (2,8).

The cosine theorem is really good, and you can use Cauchy's inequality for this problem.

Let ad=db=x, ac=2x, bc=y

then find the maximum value of y+4x.

Because the CD is the midline.

So cd 2 = 1 2(2x) 2+1 2y 2-1 4(2x) 2(y+4x) 2 (y 2+2x 2)(1+8) while cd 2=x 2+y 2 2

So 8=2x 2+y 2

So y+4x (9*8)=6 2

4 or 4 3 or 4 3 3 supplement:

a is the apex angle. A is the bottom angle and B is the top angle.

A is the bottom angle, and b is the bottom angle.

These three cases.

4 or 4 times the root number 3 or four-thirds of the root number 3.

Let bc=aab=ac=b

s abc=1 2absin acb=1 2b sin bacsin bac maximums are 1 at bac=90°, and at bac=90°, ab=b=6 5

So the triangle area is maximum.

s△abc=1/2×36/5×1=18/5

Let the waist length be x, then 5 2+(x-1) 2=x 2 is solved to x=13

So the waist length is 13

If you don't understand, please ask! Thank you!

Without losing generality, let ab ac x, then: bc 2x 3 2, bc 3 2 2x.

The formula for calculating the length of the midline of a triangle is:

cd＝（1/2）√（2ac^2＋2bc^2－ab^2）＝（1/2）√［x^2＋2（3√2－2x）^2］。

Obviously, when x 2 2 (3 2 2x) 2 achieves the minimum, cd obtains the minimum.

Ling y x 2 2 (3 2 2x) 2 x 2 2 (18 12 x 2x 4x 2) 9x 2 24 2x 36

3x）^2－8√2（3x）＋32＋4＝（3x－4√2）^2＋4。

When 3x 4 2, i.e. x 4 2 3, y has a minimum value, i.e. cd has a minimum value.

At this time, y 4, at this time, cd (1 2) y (1 2) 4 1.

That is, the minimum value of the length of the cd that satisfies the condition is 1.

If you don't use that formula! You might as well set ab=ac=2x, set the angle a, and use the cosine theorem in the triangle abc and triangle acd, respectively, and use x to represent cd, but I calculated that the minimum cd is 1(You're right?) ）