An open box is made from a square piece of cardboard 20 inches on a side by cutting identical squares from the corners and turning up the sides.(a) Express the volume of the box, V , as a function of the length of the side of the square cut from each corner, x. (b) Find and interpret V (1),V (2),V (3),V (4), and V (5). What is happening to the volume of the box as the length of the side of the square cut increases? (c) Find the domain of V.
v = x(20-2x)^2
using what you know about the shape of cubic curves, clearly there is a maximum volume for 0 <= x <= 10 (this is also the domain of v)
Naturally, volume is zero when x=0 and when x=10.
A little checking of values will show that v(x) is max when x = 10/3.
Nice one!
(a) To find the volume of the box, we first need to find the dimensions of the resulting box after cutting and folding.
If we cut identical squares with side length x from each corner of the 20-inch square cardboard, the resulting box will have dimensions (20-2x) inches by (20-2x) inches by x inches.
Therefore, the volume of the box, V, can be expressed as:
V = (20-2x)(20-2x)(x)
(b) Let's evaluate V(1), V(2), V(3), V(4), and V(5) and interpret them:
- V(1):
V(1) = (20-2(1))(20-2(1))(1) = (18)(18)(1) = 324 cubic inches
Interpretation: When we cut 1-inch squares from each corner, the resulting box has a volume of 324 cubic inches.
- V(2):
V(2) = (20-2(2))(20-2(2))(2) = (16)(16)(2) = 512 cubic inches
Interpretation: When we cut 2-inch squares from each corner, the resulting box has a volume of 512 cubic inches.
- V(3):
V(3) = (20-2(3))(20-2(3))(3) = (14)(14)(3) = 588 cubic inches
Interpretation: When we cut 3-inch squares from each corner, the resulting box has a volume of 588 cubic inches.
- V(4):
V(4) = (20-2(4))(20-2(4))(4) = (12)(12)(4) = 576 cubic inches
Interpretation: When we cut 4-inch squares from each corner, the resulting box has a volume of 576 cubic inches.
- V(5):
V(5) = (20-2(5))(20-2(5))(5) = (10)(10)(5) = 500 cubic inches
Interpretation: When we cut 5-inch squares from each corner, the resulting box has a volume of 500 cubic inches.
As the length of the side of the square cut increases, the volume of the box initially increases, reaches a maximum point, and then starts to decrease.
(c) The domain of V, in this case, would be restricted to values of x that make sense in the context of the problem. Since we are cutting squares from corners, x must be less than or equal to half the side length of the original square (x ≤ 10 inches). Additionally, x should also be greater than 0 (x > 0) since we cannot cut negative or zero-sized squares. Therefore, the domain of V is 0 < x ≤ 10.
To solve this problem, we'll take it step by step.
(a) Expressing the volume of the box, V, as a function of the length of the side of the square cut from each corner, x.
To find the volume of the box, we need to first determine the dimensions of the box after the corners are cut and the sides are folded up. In this case, cutting identical squares from each corner means that the length and width of the resulting base will be reduced by 2x (twice the length of each side cut).
The length of the base will be 20 - 2x, and the width will also be 20 - 2x. The height of the box will be x, as it is the portion that is folded up.
So, the volume formula for the box is given by V = length * width * height:
V = (20 - 2x)(20 - 2x)(x)
V = (400 - 40x - 40x + 4x^2)(x)
V = (400 - 80x + 4x^2)(x)
V = 4x^3 - 80x^2 + 400x
(b) Finding and interpreting V(1), V(2), V(3), V(4), and V(5).
To find V(1), V(2), V(3), V(4), and V(5), we substitute the values of x into the volume function V(x) we derived in part (a).
V(1) = 4(1)^3 - 80(1)^2 + 400(1) = 4 - 80 + 400 = 324
V(2) = 4(2)^3 - 80(2)^2 + 400(2) = 32 - 320 + 800 = 512
V(3) = 4(3)^3 - 80(3)^2 + 400(3) = 108 - 720 + 1200 = 588
V(4) = 4(4)^3 - 80(4)^2 + 400(4) = 256 - 1280 + 1600 = 576
V(5) = 4(5)^3 - 80(5)^2 + 400(5) = 500
Interpreting these volumes:
- V(1): When each corner square has a side length of 1 inch, the volume of the resulting box is 324 cubic inches.
- V(2): When each corner square has a side length of 2 inches, the volume of the resulting box is 512 cubic inches.
- V(3): When each corner square has a side length of 3 inches, the volume of the resulting box is 588 cubic inches.
- V(4): When each corner square has a side length of 4 inches, the volume of the resulting box is 576 cubic inches.
- V(5): When each corner square has a side length of 5 inches, the volume of the resulting box is 500 cubic inches.
By comparing the volumes, we can observe that initially, as the length of the side cut increases, the volume of the box increases. However, after a certain point (around V(3)), the volume starts to decrease.
(c) Finding the domain of V.
The domain of V represents the values of x for which the volume function is defined. In this case, x represents the length of the side cut from each corner. Therefore, the domain of V is limited by the maximum possible value of x, which is half the length of the original cardboard square.
Since the original cardboard square is 20 inches on a side, the maximum possible value of x would be 10 inches (half of 20 inches). Therefore, the domain of V is x ≥ 0 and x ≤ 10.