the sum of the base and height of triangle is 21 cm. determine the maximum area of the triangle in square cm.?
回答 (10)
2008-03-14 5:42 pm
✔ 最佳答案
Maximum area is where base = height = 10.5 cmMax area = (10.5)² / 2 = 110.25 cm² / 2 = 55.125 cm²
2008-03-14 5:57 pm
Hello:
if we call b = base of triangle
h = height of triangle
A = area of the triangle
we can write
b + h = 21cm then b = 21cm - h
A = b . h/2 = (21cm - h) . h/2
A = (21cm . h - h ²) /2
dA/dh = (21cm - 2 h ) /2
we do this equal to 0 and solve (sorry I´m from Argentina and don´t speake a good english) we call this "derivada"
0 = 10,5 cm - h then
h = 10,5 cm and b = 21cm - h = 21cm - 10,5 cm
h = 10,5 cm
The area of triangle is
A = b . h/2 = 10,5 cm . 10,5 cm/2 = 55,125 cm²
Good luck
if we call b = base of triangle
h = height of triangle
A = area of the triangle
we can write
b + h = 21cm then b = 21cm - h
A = b . h/2 = (21cm - h) . h/2
A = (21cm . h - h ²) /2
dA/dh = (21cm - 2 h ) /2
we do this equal to 0 and solve (sorry I´m from Argentina and don´t speake a good english) we call this "derivada"
0 = 10,5 cm - h then
h = 10,5 cm and b = 21cm - h = 21cm - 10,5 cm
h = 10,5 cm
The area of triangle is
A = b . h/2 = 10,5 cm . 10,5 cm/2 = 55,125 cm²
Good luck
2008-03-14 5:50 pm
This happens when each side is 10.5 cm.
b + h = 21
h = 21 - b
We must maximize:the parabola.
1/2(b(21-b))
It faces down and the zeros are at b=21 and b=0. The vertex which is the maxima is half way between the zeros at
(21+0)/2 = 10.5
b + h = 21
h = 21 - b
We must maximize:the parabola.
1/2(b(21-b))
It faces down and the zeros are at b=21 and b=0. The vertex which is the maxima is half way between the zeros at
(21+0)/2 = 10.5
2008-03-14 5:49 pm
(21/2)^2/2
= 10.5^2/2
= 110.25/2
= 55.125 square centimeters
= 10.5^2/2
= 110.25/2
= 55.125 square centimeters
2008-03-14 5:49 pm
b+h=21
h=21-b
Area=(1/2)bh=(1/2)b(21-b)
A=21b/2-b^2/2
dA/db=21/2-2b/2=0
b=21/2=10.5
h=21-10.5=10.5
d^2A/db^2 =-1 <0 which means A is maximum at b=10.5
Maximum area = (1/2)(10.5)(10.5)=55.125 sq cm
h=21-b
Area=(1/2)bh=(1/2)b(21-b)
A=21b/2-b^2/2
dA/db=21/2-2b/2=0
b=21/2=10.5
h=21-10.5=10.5
d^2A/db^2 =-1 <0 which means A is maximum at b=10.5
Maximum area = (1/2)(10.5)(10.5)=55.125 sq cm
2008-03-14 5:47 pm
b + h =21
h=21-b
A=1/2bh
A=1/2b(21-b)
A=10.5b-b^2/2
Derivative,
A'=10.5-b
We know,
A'=0
b=10.5 cm
10.5+h=21
h=10.5 cm
A=1/2*10.5*10.5= 55.125 cm^2
h=21-b
A=1/2bh
A=1/2b(21-b)
A=10.5b-b^2/2
Derivative,
A'=10.5-b
We know,
A'=0
b=10.5 cm
10.5+h=21
h=10.5 cm
A=1/2*10.5*10.5= 55.125 cm^2
2008-03-14 5:47 pm
b + h = 21
product of b and h is maximum when values are 10.5 and 10.5
area = 1/2 × 10.5 × 10.5
= 55.125 cm²
product of b and h is maximum when values are 10.5 and 10.5
area = 1/2 × 10.5 × 10.5
= 55.125 cm²
2008-03-14 5:46 pm
b + h = 21
area = (1/2)bh
since b = 21 - h
area = (1/2)(21-h)(h)
area = (1/2)(21h - h*h)
area = 10.5h - (1/2)h*h
area = maxvalue when its derivative = 0
derivative(area) = 10.5 - h
which = 0 when h = 10.5
So, b = 10.5, h = 10.5
area = (1/2)bh
since b = 21 - h
area = (1/2)(21-h)(h)
area = (1/2)(21h - h*h)
area = 10.5h - (1/2)h*h
area = maxvalue when its derivative = 0
derivative(area) = 10.5 - h
which = 0 when h = 10.5
So, b = 10.5, h = 10.5
2008-03-14 5:43 pm
Okay so b + h = 21cm
Area of a triangle = (0.5)(b)(h)
Just take some guesses as the b and h.
1,20 = 20
2, 19= 38
3, 18= 54
4, 17
5, 16
6, 15
7, 14
8, 13
9, 12
10, 11
I don't have to continue. I see the pattern. Therefore, I know that the dimension 10 and 11 will give the largest area.
It doesn't matter what is the height and what is the base.
Therefore, the area is
b x h x 0.5
10 x 11 x 0.5
= 55 cm sqaured
If this takes to long, you can use the max area formula:
Maximum area is where base = height = 10.5 cm (You get this by doing 21 divided by 2)
Max area = (10.5)² / 2 = 110.25 cm² / 2 = 55.125 cm²
Area of a triangle = (0.5)(b)(h)
Just take some guesses as the b and h.
1,20 = 20
2, 19= 38
3, 18= 54
4, 17
5, 16
6, 15
7, 14
8, 13
9, 12
10, 11
I don't have to continue. I see the pattern. Therefore, I know that the dimension 10 and 11 will give the largest area.
It doesn't matter what is the height and what is the base.
Therefore, the area is
b x h x 0.5
10 x 11 x 0.5
= 55 cm sqaured
If this takes to long, you can use the max area formula:
Maximum area is where base = height = 10.5 cm (You get this by doing 21 divided by 2)
Max area = (10.5)² / 2 = 110.25 cm² / 2 = 55.125 cm²
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