<span>Maximum area = sqrt(3)/8
Let's first express the width of the triangle as a function of it's height.
If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have
w = 1 - 2b
b = h/sqrt(3)
So
w = 1 - 2*h/sqrt(3)
The area of the rectangle is
A = hw
A = h(1 - 2*h/sqrt(3))
A = h*1 - h*2*h/sqrt(3)
A = h - 2h^2/sqrt(3)
We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0.
We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3).
The midpoint is
(0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3)
So the desired height is 0.75/sqrt(3).
Now let's calculate the width:
w = 1 - 2*h/sqrt(3)
w = 1 - 2* 0.75/sqrt(3) /sqrt(3)
w = 1 - 2* 0.75/3
w = 1 - 1.5/3
w = 1 - 0.5
w = 0.5
The area is
A = hw
A = 0.75/sqrt(3) * 0.5
A = 0.375/sqrt(3)
Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens.
A = h - 2h^2/sqrt(3)
A' = 1h^0 - 4h/sqrt(3)
A' = 1 - 4h/sqrt(3)
Now solve for 0.
A' = 1 - 4h/sqrt(3)
0 = 1 - 4h/sqrt(3)
4h/sqrt(3) = 1
4h = sqrt(3)
h = sqrt(3)/4
w = 1 - 2*(sqrt(3)/4)/sqrt(3)
w = 1 - 2/4
w = 1 -1/2
w = 1/2
A = wh
A = 1/2 * sqrt(3)/4
A = sqrt(3)/8
And the other method got us 0.375/sqrt(3). Are they the same? Let's see.
0.375/sqrt(3)
Multiply top and bottom by sqrt(3)
0.375*sqrt(3)/3
Multiply top and bottom by 8
3*sqrt(3)/24
Divide top and bottom by 3
sqrt(3)/8
Yep, they're the same.
And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
Answer:
CORRECTED QUESTION:
Two cities have nearly the same north-south line of 110 degrees Upper W. The latitude of the first city is 23 degrees Upper N, and the latitude of the second city is 36 degrees N. Approximate the distance between the cities if the average radius of Earth is 6400 km.
ANSWER: 1452.11 km
Step-by-step explanation:
Since the two cities both lies on the Northern latitude of the sphere along the same longitude, we are going to subtract the angles the latitude that each city subtend at the equator.
36 - 23 = 13 degrees i.e the angles between the with two cities on a cross section the large circle formed by the longitude and its center.
Applying the formula for the length of an arc on a sector on the large circle
(∅/ 360) x 2πR
where, ∅ = is the angle between the two cities
R = radius of the Earth.
13/360 x 2 x π x 6400 = 1452.11 km
Answer:
f = 26
d=112
e=42
Step-by-step explanation:
as far as what i can see, the first given angle is 112 and the second angle is 42.
These two angles are the angles on one straightline, therefore the missing angle will be 180 - 112 - 42 = 26 (Supplementary angle). This angle is not required and it is the missing angle on the left hand side. However, this angle should be equal to that of f (Opposite angle).
By the same way, d and e are the two opposite angles for 112 and 42, respectively.
The rate of change is +2 degrees per hour (an increase of 2 degrees per hour)
Answer:
4x+4
Step-by-step explanation:
