an open box is to be made by cutting small congruent squares from corners of a 12cm by 12cm . sheet of tin and bending up the sides . how large should the squares cut from the corners to be make the box hold as much as possible ?
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To find the size of the squares that should be cut from the corners in order to make a box that can hold as much as possible, we need to consider the volume of the box.
Let's start by visualizing the scenario. We have a 12cm by 12cm sheet of tin, and we will cut small congruent squares from each corner. By folding up the sides, we can form an open box with no top.
To determine the dimensions of the box, we need to find the length, width, and height of the box.
Cutting squares from each corner of the tin sheet will reduce both the length and the width of the box. Let's assume we cut squares of side "x" from each corner.
After cutting the squares, the length of the box will be (12 - 2x) cm, and the width will also be (12 - 2x) cm.
The height of the box will be equal to the side length of the squares we cut, which is "x" cm.
Now, we can calculate the volume of the box. The volume of a rectangular prism is given by the formula: V = length * width * height.
Therefore, the volume of the box is:
V = (12 - 2x) * (12 - 2x) * x
V = x(144 - 24x + 4x^2)
V = 4x^3 - 24x^2 + 144x
To find the maximum volume, we need to find the critical points of the volume function. We can do this by taking the derivative of the volume function and setting it equal to zero.
dV/dx = 12x^2 - 48x + 144 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
x = (-(-48) ± √((-48)^2 - 4 * 12 * 144)) / (2 * 12)
x = (48 ± √(2304 - 6912)) / 24
x = (48 ± √(-4608)) / 24
Since the square root of a negative number is not a real number, we conclude that there are no real solutions for x. Hence, our quadratic equation does not have any critical points.
Therefore, we need to consider the endpoints of the interval. x must be greater than zero, but it should also be less than half the length of the side of the tin sheet.
In this case, the side length of the tin sheet is 12 cm. So, the interval for x is 0 < x < 6.
Since there are no critical points, we can evaluate the volume of the box at the endpoints of the interval to find the maximum volume.
When x = 0, the volume of the box is:
V = 4(0^3) - 24(0^2) + 144(0)
V = 0
When x = 6, the volume of the box is:
V = 4(6^3) - 24(6^2) + 144(6)
V = 864 - 864 + 864
V = 864
Therefore, the maximum volume of the box is 864 cm^3.
To achieve this maximum volume, we need to cut squares with side length 6 cm from each corner of the tin sheet.