mat e. matics starts to construct a series of triangles. all the triangles are drawn such that the sum of the base and height are always 10cm. determine the maximum area of all these triangles.
area=1/2 b*h
I suspect the maximum will occur then base=height.
I need solution A quadratic functions completing the square a player bumps a volleyball with an initial vertical velocity of 20 ft/s
1/Write a functions h in standard form for the boll`s height in feet in terms of the time t in seconds after the ball is hit
2/Complete the square to rewrite h in vertex form
3/what is the maximum height of the ball?
4/What if...?Suppose the volleyball were hit under the same conditions, but with an initial velocity of 32 ft/s .How much higher would the ball go?
To determine the maximum area of the triangles, we can use the formula for the area of a triangle: A = (1/2) * base * height.
In this case, we are given that the sum of the base and height is always 10 cm. Let's assume the length of the base is x cm. Since the sum of the base and height is 10 cm, we can express the height as (10 - x) cm.
Now we can calculate the area of the triangle in terms of x:
A = (1/2) * x * (10 - x)
To find the maximum area, we need to find the value of x that maximizes this equation.
To do this, we can use calculus. Let's differentiate the equation A with respect to x and find its critical points:
dA/dx = (1/2) * (10 - 2x)
Setting dA/dx equal to zero:
(1/2) * (10 - 2x) = 0
10 - 2x = 0
2x = 10
x = 5
So, the critical point is x = 5.
Now, we need to determine whether this critical point is a maximum or a minimum. We can check the second derivative to determine this:
d^2A/dx^2 = -2
Since the second derivative is negative, we can conclude that x = 5 corresponds to a maximum area.
To find the maximum area, we substitute the value of x back into the area equation:
A = (1/2) * 5 * (10 - 5)
A = (1/2) * 5 * 5
A = 12.5 cm^2
Therefore, the maximum area of all the triangles is 12.5 square centimeters.