<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:
the nearest thousand is 15 000.00
<span>You are given a car moved at a constant velocity during the first hour. It stopped for 2 hours at a mall and then moved ahead again at a constant velocity for the next 3 hours. Then you are given the car that has finally returned to its starting point with a constant velocity in the next 2.5 hours. The graph that best represents the car's motion is First straight line joins ordered pairs 0, 0 and 1, 60, second straight line joins 1, 60 and 3, 60, third straight line joins 3, 60 and 6, 100 and fourth straight line joins 6, 100 and 8.5, 0.</span>
First we solve what we can solve.
<span>y</span>-3= 2/3<span>(</span>x-1)
We first multiply
<span>y</span>-3= 2/3 (x) - 2/3
Then we move the -3 and it becomes +3 on the other side
y= 2/3 (x) - 2/3 + 3
And we solve what we can to get our answer.
y= 2/3 (x) + 2 1/3
Given :
Natalie is going mountain biking. She can buy a bike for $250 or she can rent a bike for $30 an hour.
In both cases, she must also rent a helmet for $5 an hour.
To Find :
Which inequality shows the number of hours Natalie must bike for the cost of buying a bike to be less than renting a bike.
Solution :
Let, after t hours total money required is ( if she rent bike ).
T = 30t.
Money required to purchase bike , M = $250.
For cost of buying a bike to be less than renting a bike :
Hence, this is the required solution.
