Answer:
Connect the intersections of the diameters and the circle with a segment
Step-by-step explanation:
Assuming the construction creates two orthogonal diameters, their ends will be the vertices of the inscribed square. Connecting them with segments in order around the circle will create the square.
okay well usually what I would tell you would be to use the ABC method. you would multiply A (2) by C (6) to get twelve. so you need two numbers that multiply to twelve and add to -4 (the B value). but there are no numbers that fit this description so you have to use the GCF method instead
for this method you must find the GCF for all three numbers which in this case is 2. 2(x^2-2x+3) all these numbers are now prime so this is the factored form
The questions for this problem would be:
1. What is the dimensions of the box that has the maximum volume?
2. What is the maximum volume of the box?
Volume of a rectangular box = length x width x height
From the problem statement,
length = 12 - 2x
width = 9 - 2x
height = x
where x is the height of the box or the side of the equal squares from each corner and turning up the sides
V = (12-2x) (9-2x) (x)
V = (12 - 2x) (9x - 2x^2)
V = 108x - 24x^2 -18x^2 + 4x^3
V = 4x^3 - 42x^2 + 108x
To maximize the volume, we differentiate the expression of the volume and equate it to zero.
V = 4x^3 - 42x^2 + 108x
dV/dx = 12x^2 - 84x + 108
12x^2 - 84x + 108 = 0x^2 - 7x + 9 = 0
Solving for x,
x1 = 5.30 ; Volume = -11.872 (cannot be negative)
x2 = 1.70 ; Volume = 81.872
So, the answers are as follows:
1. What is the dimensions of the box that has the maximum volume?
length = 12 - 2x = 8.60
width = 9 - 2x = 5.60
height = x = 1.70
2. What is the maximum volume of the box?
Volume = 81.872
For many things like putting flooring in or like calculating how much liquid can pass threw a pipe
