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Sagot :
To solve the problem of finding the minimum surface area for a rectangular box with a square base and an open top, where the volume is fixed at [tex]\(24 \, \text{cm}^3\)[/tex], we need to follow several steps.
### Step-by-Step Solution
1. Express the volume constraint:
The volume [tex]\(V\)[/tex] of the box is given by:
[tex]\[ V = x^2h \][/tex]
Where [tex]\(x\)[/tex] is the side of the square base and [tex]\(h\)[/tex] is the height of the box. We are given that:
[tex]\[ x^2h = 24 \][/tex]
2. Express [tex]\(h\)[/tex] in terms of [tex]\(x\)[/tex]:
Solve for [tex]\(h\)[/tex]:
[tex]\[ h = \frac{24}{x^2} \][/tex]
3. Express the surface area:
The surface area [tex]\(S\)[/tex] consists of the base ([tex]\(x^2\)[/tex]) and the four sides ([tex]\(4xh\)[/tex]), but does not include the top since it is open. So,
[tex]\[ S = x^2 + 4xh \][/tex]
Substitute [tex]\(h\)[/tex] from the volume constraint:
[tex]\[ S = x^2 + 4x \left(\frac{24}{x^2}\right) \][/tex]
Simplify the surface area function:
[tex]\[ S = x^2 + \frac{96}{x} \][/tex]
4. Find the critical points by differentiating:
To minimize [tex]\(S\)[/tex], find its derivative with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{dS}{dx} = 2x - \frac{96}{x^2} \][/tex]
5. Set the derivative to zero to find critical points:
[tex]\[ 2x - \frac{96}{x^2} = 0 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 2x = \frac{96}{x^2} \][/tex]
[tex]\[ 2x^3 = 96 \][/tex]
[tex]\[ x^3 = 48 \][/tex]
[tex]\[ x = \sqrt[3]{48} = 2 \sqrt[3]{6} \][/tex]
6. Find the height [tex]\(h\)[/tex]:
Substitute the value of [tex]\(x\)[/tex] back into the volume equation:
[tex]\[ h = \frac{24}{x^2} \][/tex]
With [tex]\(x = 2 \sqrt[3]{6}\)[/tex]:
[tex]\[ h = \frac{24}{(2 \sqrt[3]{6})^2} \][/tex]
Simplify:
[tex]\[ h = \frac{24}{4 \sqrt[3]{36}} \][/tex]
[tex]\[ h = \frac{24}{4 \cdot 6^{2/3}} \][/tex]
[tex]\[ h = \frac{24}{4 \cdot 6^{2/3}} = \frac{6}{6^{2/3}} = \sqrt[3]{6} \][/tex]
7. Calculate the minimum surface area [tex]\(S\)[/tex]:
Substitute [tex]\(x = 2 \sqrt[3]{6}\)[/tex] and [tex]\(h = \sqrt[3]{6}\)[/tex] into the surface area function:
[tex]\[ S = (2 \sqrt[3]{6})^2 + 4(2 \sqrt[3]{6})(\sqrt[3]{6}) \][/tex]
Simplify the surface area expression:
[tex]\[ S = 4 \cdot 6^{2/3} + 8 \cdot 6^{2/3} \][/tex]
[tex]\[ S = 12 \cdot 6^{2/3} \][/tex]
Thus, the minimum surface area needed to construct the box is:
[tex]\[ \boxed{12 \cdot 6^{2/3}} \quad \text{cm}^2 \][/tex]
### Step-by-Step Solution
1. Express the volume constraint:
The volume [tex]\(V\)[/tex] of the box is given by:
[tex]\[ V = x^2h \][/tex]
Where [tex]\(x\)[/tex] is the side of the square base and [tex]\(h\)[/tex] is the height of the box. We are given that:
[tex]\[ x^2h = 24 \][/tex]
2. Express [tex]\(h\)[/tex] in terms of [tex]\(x\)[/tex]:
Solve for [tex]\(h\)[/tex]:
[tex]\[ h = \frac{24}{x^2} \][/tex]
3. Express the surface area:
The surface area [tex]\(S\)[/tex] consists of the base ([tex]\(x^2\)[/tex]) and the four sides ([tex]\(4xh\)[/tex]), but does not include the top since it is open. So,
[tex]\[ S = x^2 + 4xh \][/tex]
Substitute [tex]\(h\)[/tex] from the volume constraint:
[tex]\[ S = x^2 + 4x \left(\frac{24}{x^2}\right) \][/tex]
Simplify the surface area function:
[tex]\[ S = x^2 + \frac{96}{x} \][/tex]
4. Find the critical points by differentiating:
To minimize [tex]\(S\)[/tex], find its derivative with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{dS}{dx} = 2x - \frac{96}{x^2} \][/tex]
5. Set the derivative to zero to find critical points:
[tex]\[ 2x - \frac{96}{x^2} = 0 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 2x = \frac{96}{x^2} \][/tex]
[tex]\[ 2x^3 = 96 \][/tex]
[tex]\[ x^3 = 48 \][/tex]
[tex]\[ x = \sqrt[3]{48} = 2 \sqrt[3]{6} \][/tex]
6. Find the height [tex]\(h\)[/tex]:
Substitute the value of [tex]\(x\)[/tex] back into the volume equation:
[tex]\[ h = \frac{24}{x^2} \][/tex]
With [tex]\(x = 2 \sqrt[3]{6}\)[/tex]:
[tex]\[ h = \frac{24}{(2 \sqrt[3]{6})^2} \][/tex]
Simplify:
[tex]\[ h = \frac{24}{4 \sqrt[3]{36}} \][/tex]
[tex]\[ h = \frac{24}{4 \cdot 6^{2/3}} \][/tex]
[tex]\[ h = \frac{24}{4 \cdot 6^{2/3}} = \frac{6}{6^{2/3}} = \sqrt[3]{6} \][/tex]
7. Calculate the minimum surface area [tex]\(S\)[/tex]:
Substitute [tex]\(x = 2 \sqrt[3]{6}\)[/tex] and [tex]\(h = \sqrt[3]{6}\)[/tex] into the surface area function:
[tex]\[ S = (2 \sqrt[3]{6})^2 + 4(2 \sqrt[3]{6})(\sqrt[3]{6}) \][/tex]
Simplify the surface area expression:
[tex]\[ S = 4 \cdot 6^{2/3} + 8 \cdot 6^{2/3} \][/tex]
[tex]\[ S = 12 \cdot 6^{2/3} \][/tex]
Thus, the minimum surface area needed to construct the box is:
[tex]\[ \boxed{12 \cdot 6^{2/3}} \quad \text{cm}^2 \][/tex]
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