A square sheet of cardboard has sides 18cm. Four equal portions are cut away from the corner and discard. The sides are then twiced up to form a restangular box. What are the dimension of the box formed that will have max volume?
To find the dimensions of the box that will have the maximum volume, we can follow these steps:
1. Visualize the problem:
Imagine a square sheet of cardboard with sides of 18 cm. The four corners are cut away, forming four equal smaller squares, which are discarded. Finally, the sides of the remaining shape (the cardboard with the corners cut off) are folded up to form a rectangular box.
2. Set up the variables:
Let's assume that the side length of each smaller square cut away is "x." We want to find the dimensions of the rectangular box, so let's call the length of the rectangular box "l," and the width "w."
3. Calculate the dimensions of the box:
When the corners are cut away, we remove 4 squares with sides of length "x." Therefore, the length (l) of the rectangular box will be the original side length (18 cm) minus twice the length of each small square (2x). So, l = 18 cm - 2x.
Similarly, the width (w) of the rectangular box will be the original side length (18 cm) minus twice the width of each small square. Since all sides of the original square are equal, the width will also be (18 cm - 2x).
Therefore, the dimensions of the rectangular box are:
Length (l) = 18 cm - 2x
Width (w) = 18 cm - 2x
4. Determine the volume:
The volume (V) of a rectangular box is calculated by multiplying its length, width, and height. In this case, the height will be equal to the length of the small squares that were cut away, which is "x."
Therefore, the volume (V) of the box is: V = l * w * x.
5. Find the maximum volume:
To find the maximum volume, we need to maximize the function V = l * w * x. Replace l and w with their respective values:
V = (18 cm - 2x) * (18 cm - 2x) * x
Expand and simplify the equation to get a quadratic equation in terms of x.
6. Maximize the volume:
To find the maximum volume, we can take the derivative of the above equation with respect to x, set it equal to zero, and solve for x. The value of x will give us the dimensions of the box that maximize the volume.
Once we find the value of x, substitute it back into the expressions for l and w to find the dimensions of the box.
Therefore, following these steps, you can find the dimensions of the rectangular box that will have the maximum volume from the given square sheet of cardboard.
L = 18-2x
w = 18-2x
h = x
V = x(18-2x)^2
dV/dx = 0 for max or min
0 = x (2)(18-2x)(-2) + 1 (18-2x)^2
0 = -4x (18-2x) + 324 - 72 x + 4 x^2
0 = -72 x + 8 x^2 + 324 -72 x + 4 x^2
0 = 12 x^2 - 144 x + 324
0 = x^2 - 12 x + 27
0 = (x-3)(x-9)
x = 3 or 9
but 9 will not work (is V=0) so use 3