A math problem to find the maximum area of a triangle

Let bc=x

ac=√2x

cosc=(x^2+2x^2-4)/(2√2x^2)=(3x^2-4)/(2√2x^2)

SinC = 1-[(3x 2-4) 2 (2 2x 2) 2] (x 4+24x 2-16) (2 2x 2) Area of the triangle ABC = 1 2BC*ac*sinc== (-x 4+24x 2-16) 4

-x^2-12)^2+128]/2

So when x 2 = 12, that is, x = 2 3, the maximum value of the area of the area of the largest triangle abc ( 128) 4 = 2 2 I rely on, I miscalculated 2 times in a row, and spent 2 pieces of manuscript paper, and the landlord gave me + points.

With AB as the bottom edge, the area should be the largest, that is, the height should be the largest, and the area should be the largest when the BC is high according to the nature of the triangle. The area is 2

The range and maxima problem in triangles is a problem that students are afraid of in the process of learning to solve triangles, which not only requires the use of trigonometric transformations and the sine and cosine theorem, but also often involves fundamental inequalities and finding the range of functions. In the college entrance examination, there are various types of questions, such as multiple-choice questions, fill-in-the-blank questions and answer questions, and the difficulty of the test questions belongs to the middle and high-end questions.

Usage scenario: In a general triangle.

Problem Solving Template: Step 1 Decide on the appropriate formula through observation and analysis;

The second step is to solve the problem by using the induction formula of trigonometric functions, identity transformation, corner transformation, sine and cosine theorem, etc., through operation and deformation, and the problem is transformed into trigonometric transformation, shouting mega beat, basically not Zheng Xian equation, function value range and other types.

Step 3 Draw conclusions.

Example] Find the maximum value of the area satisfying , .

Analysis] Let , then , according to the area formula

From the cosine theorem, substitution is obtained, and the trilateral relationship of the triangle has and , so that when (i.e., time), the maximum value is obtained

Summary] Combined with the knowledge of functions, this question takes the triangle that students are familiar with as the carrier, and examines the knowledge of area formula, cosine theorem and other knowledge, which is a good problem to investigate and solve triangles.

Let the two sides of the Sanyou reed angle be b,c, and their first and remaining angles are good celery bands a,s abc=(1 2)b*c*sina, when a=90 degrees, sina has a maximum value of 1, so the maximum area is bc 2

When b= c, the triangle area is maximum:

s=a^2sinbsinc/(2sina)=18(sinb)^2/sina

It is known again by: tana = 3 4

Get: sina = 3 5, cosa = 4 5

sinb=sin(90°-a2)=cos(a2) is obtained by the half-angle formula: (sinb) 2=(1+cosa) 2=9 10, so the maximum area is: s=18*(9 10) (3 5)=9*3=27

(b-c)>=0

b^2+c^2>=2bc

From this inequality, the maximum value of BC is calculated.

The bottom of the BCD is fixed, and the area is the largest when its high DE is maximum. Point A moves on a circle with C as the center and radius 1. Let bca= , abc= , and use the cosine theorem in abc:

ab = 1 +2 -2 1 2cos = 5-4cos sine theorem:

sinβ=sinα/√（5-4cosα）

cosβ=√[1-sin²α/（5-4cosα）]=√（5-4cosα-sin²α）/√（5-4cosα）=√（4-4cosα+cos²α）/√（5-4cosα）=（2-cosα）/√（5-4cosα）

de=absin（60°+β=√（5-4cosα）sin（60°+β

（5-4cosα）[sin60°cosβ+cos60°sinβ]=√（5-4cosα）[3/

1/2）[√3（2-cosα）+sinα]=（1/2）[2√3-√3cosα+sinα]=√3+（1/

3+（cos60°sinα-sin60°.cosα）=√3+sin（α-60°）

At 60°=90°, =150°, the area is maximum:

demax=1+3, area=(1+3)2 2=1+3

When b= c, the triangle area is maximum:

s=a^2sinbsinc/(2sina)=18(sinb)^2/sina

It is known again by: tana = 3 4

Get: sina = 3 5, cosa = 4 5

sinb=sin(90°-a2)=cos(a2) is obtained by the half-angle formula: (sinb) 2=(1+cosa) 2=9 10, so the maximum area is: s=18*(9 10) (3 5)=9*3=27

When b= c, the triangle area is maximum:

s=a^2sinbsinc/(2sina)=18(sinb)^2/sina

It is known again by: tana = 3 4

Get: sina = 3 5, cosa = 4 5

sinb=sin(90°-a2)=cos(a2) is obtained by the half-angle formula: (sinb) 2=(1+cosa) 2=9 10, so the maximum area is: s=18*(9 10) (3 5)=9*3=27

The largest area of the triangle is an equilateral triangle, and the area of an equilateral triangle is the largest area of the triangle in the same circumference. The circumference is known, and the area of the equilateral triangle is 1 9d sin60°

Let the side lengths of the triangle be a, b, and c. respectivelywhere c is the hypotenuse.

Known: a+b+c=l

Pythagorean theorem a 2 + b 2 = c 2

Area s=1 2*a*b

To maximize s, you need to make a b a maximum.

Because A2+B 2>=2AB, when A=B (when it is an isosceles right triangle), AB takes the maximum value, i.e., AB (A 2+B 2) 2=C 2 2

So maxs=c 2 4

And because a+b+c=l,a=b,c=root number 2 a, collated, c=2, then the maximum value of s is l 2 (12+8*2

Note: c 2 means c to the power of 2, and 2 means to the power of 2, which means the root number 2.

The answer is: triangles.

The area of ABC is maximum 12 5

The answer is as follows: C=2, B=2A, Cosa=(B2+C2-A2) 2BC=(5A2-4) 8A, Sina= (1-Cos 2A)= 25A4+104A2-16) 8A, S-ABC Area=1 2*BC*Sina

2a*sina

-25a 4+104a 2-16) 4, ling, s-abc area = s, then there is.

s= (25a 4+104a 2-16) 4, squared on both sides.

25a 4+104a 2-16(1+s 2)=0, to make the equation have a solution, 0, that is, there is.

104) 2-4*(-25)*[16(1+s 2)] 0,13 2 25(1+s 2),s 2 144 25,s 12 5, then, the maximum area of the triangle abc is: 12 5,

a+b=4, ab=m, 5x 2-6x-8=0, i.e. (x-2)(5x+4)=0, so x=2 or -4 5, i.e. cosc=2 or -4 5, because -1 cosc 1, so cosc=-4 5, so sinc=3 5

Because a+b 2 ab, so 4 2 m, so m 4, i.e. ab 4 so s=(1 2) ab sinc=(1 2) ab (3 5) (1 2) 4 (3 5)=6 5

That is, the maximum value of s is 6 5