<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>
Both distances 3966 and 4746 were done in 6 hours, so the speeds are:
speed against wind = 3966/6=661 km/h (difference of airplane and wind)
speed with wind = 4746/6=791 km/h (sum of airplane and wind)
The speed of airplane is greater than that of the wind, so this is a sum and difference problem.
Greater speed (plane) = (sum+difference)/2=(791+661)/2=726 km/h
Lesser speed (wind) = (sum-difference)/2 = (791-661)/2 = 65 km/h
First, let’s find the area of the larger square. We know that the smaller square extend out 2ft and is2ft up. Well take these measures, and subtract the 2ft from the 12 ft at the top, and get 10ft, then add the 2ft to the left side, and find that it’s 6ft. We have these measures: 6ft 6ft 10ft 10ft for the larger rectangle. Let’s find the area. The are formula for a rectangle is Base(10ft) times height(6ft) . Thus 60ftsqrd. Next, let’s do the smaller rectangle. We do the same formula, 4 times 2, and get 8ftsqrd, add them together: 60ft + 8ft = 68ft!
One number is 26 the other is 23
x+y=49
Y=x+3,X (now substitute it in the equation)
x+3+x=49
2x+3=49
2x=46. X=23
Y=x+3. Y=23+3 =26 for checking 23+26=49
