Hi, Please help me find: V, a, and b. Ty
An open box is to be made from a flat piece of material 8 inches long and 2 inches wide by cutting equal squares of length x from the corners and folding up the sides.
Write the volume V of the box as a function of x. Leave it as a product of factors, do not multiply out the factors.
V= ?
If we write the domain of the volume of the box as an open interval in the form (a,b), then what is a=?
a= ?
and what is b=?
b=
MMMhhh?
Both I and bobpursley answered you .
http://www.jiskha.com/display.cgi?id=1390567139
Why are you reposting?
L = 8 - 2x
W = 2 - 2x
V = L W x
V = (8-2x)(2-2x)x
V = 4(4-x)(1-x)x
V = 4 x (4 -5x + x^2)
V/4 = x^3 - 5x^2 + 4x
If x is > 1 that is impossible because the width will be negative
if x is 0, our volume is 0
I wonder where V is maximum?
where is derivative = 0?
f' = 3 x^2 -10 x + 4
that is 0 when
x = [ 10 +/- sqrt(100 - 48)]/6
x = [ 10 +/- 7.21]/6
x = .465 or 2.88
2.88 is outside of our domain or x so
max V when x = .465
so domain of V is from 0 to V(.465)=3.52
by the way, I would call that the range of V
the domain of x is 0 to 1
To find the volume V of the box, we need to determine the dimensions of the box after cutting the squares and folding up the sides.
The length of the box would be (8 - 2x) inches, calculated by subtracting twice the value of x from the original length.
Similarly, the width of the box would be (2 - 2x) inches, obtained by subtracting twice the value of x from the original width.
The height of the box, which is the length of the square cut from the corners, would be x inches.
To find the volume V, we multiply the length, width, and height together:
V = (8 - 2x) * (2 - 2x) * x
Now, let's simplify this expression by leaving it as a product of factors without multiplying them out:
V = x(8 - 2x)(2 - 2x).
This is the volume V of the box as a function of x.
Now, to determine the domain of the volume of the box, we need to consider the possible values of x.
Since we are cutting squares from the corners, the length of each side of these squares cannot exceed half the length or width of the original piece of material. Therefore, x ≤ min(8/2, 2/2) = min(4, 1) = 1.
So, the domain of the volume of the box is (0, 1) as we cannot have the length of the squares extend beyond the limits of the material.
Therefore, a = 0 and b = 1.