Square side of x are cut from each corner of an 8 in x 5 in rectangle (see figures), so that its can be folded to make a box no top, define a function in terms of x that can represent the volume of this box
Answers
Step-by-step explanation:
This is a classical problem which has delighted calculus students down through the ages.
We first express the volume of the box as a function of x, the side of the square which is cut out of each corner.
The resulting box will have a base which is a square which all of whose sides will have length 12 - 2x. The height of the box will be x. Therefore, the volume will be
V = (12 - 2x)2x
= 144x - 48x2 + 4x3
Since the amount you cut out can't be negative, and since you can't cut out more than half of the side, the domain of the function is
0 < x < 6
If we differentiate the volume and set it equal to zero, we get
V' =12x2 - 96x + 144 = 0
This gives us a quadratic to solve. Factor
12(x2 - 8x + 12) = 0
12(x - 6)(x - 2) = 0
Set the factors = 0.
x - 6 = 0
x = 6
x - 2 = 0
x = 2
This gives us two points to check
V(6) = 144(6) - 48(6)2 + 4(6)3
= 144(6) - 48(36) + 4(216)
= 864 - 1728 + 864
= 0
This is probably not the maximum volume. Let's try x = 2
V(2) = 144(2) - 48(2)2 + 4(2)3
= 144(2) - 48(4) + 4(8)
= 288 - 192 + 32
= 128
Answer:
length 12 - 2x. The height of the box will be x. Therefore, the volume will be
V = (12 - 2x)2x
= 144x - 48x2 + 4x3
Since the amount you cut out can't be negative, and since you can't cut out more than half of the side, the domain of the function is
0 < x < 6
If we differentiate the volume and set it equal to zero, we get
V' =12x2 - 96x + 144 = 0
This gives us a quadratic to solve. Factor
12(x2 - 8x + 12) = 0
12(x - 6)(x - 2) = 0
Set the factors = 0.
x - 6 = 0
x = 6
x - 2 = 0
x = 2
This gives us two points to check
V(6) = 144(6) - 48(6)2 + 4(6)3
= 144(6) - 48(36) + 4(216)
= 864 - 1728 + 864
= 0
This is probably not the maximum volume. Let's try x = 2
V(2) = 144(2) - 48(2)2 + 4(2)3
= 144(2) - 48(4) + 4(8)
= 288 - 192 + 32
= 128V(0) = 144(0) - 48(0)2 +4(0)3
= 0
We conclude that the minimum possible volume is 0 which happens at the end points, and the maximum possible volume is 128 when x = 2.
The two situations at the endpoints of the domain are degenerate cases. When x = 0, you just have your original sheet of cardboard from which you have not yet cut out the corners let alone fold up the edges. If x =