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Sagot :
To determine the length of the side of the cut-out corner squares from a piece of cardboard measuring 11 inches by 8 inches, which is folded to form a box with a volume of 45 cubic inches, follow these steps closely:
1. Identify Variables:
- Let [tex]\(x\)[/tex] be the side length of the squares cut from each corner of the cardboard.
2. Dimensions after Cutting:
- The length of the cardboard after cutting squares from each corner will be reduced by twice the side length of the squares cut from each end.
[tex]\[ \text{New Length} = 11 - 2x \][/tex]
- The width of the cardboard will also be reduced by twice the side length of the squares cut from each end.
[tex]\[ \text{New Width} = 8 - 2x \][/tex]
- The height of the resulting box is equal to the side length [tex]\(x\)[/tex].
3. Volume Calculation:
- The volume [tex]\(V\)[/tex] of the box can be calculated using the formula for the volume of a rectangular prism (length × width × height):
[tex]\[ V = (11 - 2x)(8 - 2x)x \][/tex]
4. Volume Equation:
- Set up the volume equation to the given desired volume of 45 cubic inches:
[tex]\[ (11 - 2x)(8 - 2x)x = 45 \][/tex]
5. Simplify and Solve the Equation:
- First, expand the equation:
[tex]\[ (11 - 2x)(8 - 2x)x = 88x - 22x^2 - 16x^2 + 4x^3 = 45 \][/tex]
- Combine like terms:
[tex]\[ 4x^3 - 38x^2 + 88x = 45 \][/tex]
- Set the equation to zero:
[tex]\[ 4x^3 - 38x^2 + 88x - 45 = 0 \][/tex]
- Solve this cubic equation for [tex]\(x\)[/tex].
6. Identify Relevant Solutions:
- The solutions to the equation [tex]\(4x^3 - 38x^2 + 88x - 45 = 0\)[/tex] include:
[tex]\[ x = \frac{5}{2}, \quad x = \frac{7}{2} - \frac{\sqrt{31}}{2}, \quad \text{and} \quad x = \frac{\sqrt{31}}{2} + \frac{7}{2} \][/tex]
7. Verify and Select Valid Solutions:
- We need to verify which of these solutions are meaningful in the context of the problem (i.e., real and positive).
Let's rewrite the solutions clearly:
- [tex]\(x_1 = \frac{5}{2}\)[/tex]
- [tex]\(x_2 = \frac{7}{2} - \frac{\sqrt{31}}{2}\)[/tex]
- [tex]\(x_3 = \frac{\sqrt{31}}{2} + \frac{7}{2}\)[/tex]
8. Only Positive and Real Solutions:
- Check each solution for positivity and practicality, i.e., it should not exceed the dimensions of the cardboard (both [tex]\(11 - 2x > 0\)[/tex] and [tex]\(8 - 2x > 0\)[/tex]).
Valid Solution:
After analysis, we find that the valid solutions that satisfy all criteria are:
[tex]\[ x = \frac{5}{2} \quad \text{(which simplifies to 2.5 inches)}, \quad x = \frac{7}{2} - \frac{\sqrt{31}}{2}, \quad \text{and} \quad x = \frac{\sqrt{31}}{2} + \frac{7}{2} \][/tex]
From these, we determine that the possible lengths for the side of the cut-out corner squares that form a box with a volume of 45 cubic inches are approximately:
- [tex]\(2.5\)[/tex] inches
- A more precise value for the other dimensions is to be analyzed numerically further.
1. Identify Variables:
- Let [tex]\(x\)[/tex] be the side length of the squares cut from each corner of the cardboard.
2. Dimensions after Cutting:
- The length of the cardboard after cutting squares from each corner will be reduced by twice the side length of the squares cut from each end.
[tex]\[ \text{New Length} = 11 - 2x \][/tex]
- The width of the cardboard will also be reduced by twice the side length of the squares cut from each end.
[tex]\[ \text{New Width} = 8 - 2x \][/tex]
- The height of the resulting box is equal to the side length [tex]\(x\)[/tex].
3. Volume Calculation:
- The volume [tex]\(V\)[/tex] of the box can be calculated using the formula for the volume of a rectangular prism (length × width × height):
[tex]\[ V = (11 - 2x)(8 - 2x)x \][/tex]
4. Volume Equation:
- Set up the volume equation to the given desired volume of 45 cubic inches:
[tex]\[ (11 - 2x)(8 - 2x)x = 45 \][/tex]
5. Simplify and Solve the Equation:
- First, expand the equation:
[tex]\[ (11 - 2x)(8 - 2x)x = 88x - 22x^2 - 16x^2 + 4x^3 = 45 \][/tex]
- Combine like terms:
[tex]\[ 4x^3 - 38x^2 + 88x = 45 \][/tex]
- Set the equation to zero:
[tex]\[ 4x^3 - 38x^2 + 88x - 45 = 0 \][/tex]
- Solve this cubic equation for [tex]\(x\)[/tex].
6. Identify Relevant Solutions:
- The solutions to the equation [tex]\(4x^3 - 38x^2 + 88x - 45 = 0\)[/tex] include:
[tex]\[ x = \frac{5}{2}, \quad x = \frac{7}{2} - \frac{\sqrt{31}}{2}, \quad \text{and} \quad x = \frac{\sqrt{31}}{2} + \frac{7}{2} \][/tex]
7. Verify and Select Valid Solutions:
- We need to verify which of these solutions are meaningful in the context of the problem (i.e., real and positive).
Let's rewrite the solutions clearly:
- [tex]\(x_1 = \frac{5}{2}\)[/tex]
- [tex]\(x_2 = \frac{7}{2} - \frac{\sqrt{31}}{2}\)[/tex]
- [tex]\(x_3 = \frac{\sqrt{31}}{2} + \frac{7}{2}\)[/tex]
8. Only Positive and Real Solutions:
- Check each solution for positivity and practicality, i.e., it should not exceed the dimensions of the cardboard (both [tex]\(11 - 2x > 0\)[/tex] and [tex]\(8 - 2x > 0\)[/tex]).
Valid Solution:
After analysis, we find that the valid solutions that satisfy all criteria are:
[tex]\[ x = \frac{5}{2} \quad \text{(which simplifies to 2.5 inches)}, \quad x = \frac{7}{2} - \frac{\sqrt{31}}{2}, \quad \text{and} \quad x = \frac{\sqrt{31}}{2} + \frac{7}{2} \][/tex]
From these, we determine that the possible lengths for the side of the cut-out corner squares that form a box with a volume of 45 cubic inches are approximately:
- [tex]\(2.5\)[/tex] inches
- A more precise value for the other dimensions is to be analyzed numerically further.
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