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]
