Question

ex: a) squares which sides of length \\(x\\) are cut out of each corner of a rectangular piece of cardboard measuring \\(23\text{ft}\\) by \\(13\text{ft}\\). the resulting piece of cardboard is then folded into a box without a lid. find the volume of the largest box that can be formed this way.

Explanation:

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<reasoning>

Express the volume function

The dimensions of the cardboard are \(23\text{ ft}\) by \(13\text{ ft}\).
Cutting out squares of side length \(x\) from each corner yields:

• Length: \(l = 23 - 2x\)
• Width: \(w = 13 - 2x\)
• Height: \(h = x\)

The volume \(V(x)\) is:
\[
V(x) = x(23 - 2x)(13 - 2x) = 4x^3 - 72x^2 + 299x
\]
The physical domain is \(0 < x < 6.5\).

Find the derivative

Differentiate \(V(x)\) with respect to \(x\):
\[
V'(x) = 12x^2 - 144x + 299
\]

Solve for critical points

Set \(V'(x) = 0\):
\[
12x^2 - 144x + 299 = 0
\]
Using the quadratic formula:
\[
x = \frac{144 \pm \sqrt{(-144)^2 - 4(12)(299)}}{24}
\]
\[
x = \frac{144 \pm \sqrt{20736 - 14352}}{24} = \frac{144 \pm \sqrt{6384}}{24}
\]
\[
x = \frac{144 \pm 4\sqrt{399}}{24} = 6 \pm \frac{\sqrt{399}}{6}
\]
Evaluating the roots:

• \(x_1 = 6 + \frac{\sqrt{399}}{6} \approx 6 + 3.33 = 9.33\) (outside domain)
• \(x_2 = 6 - \frac{\sqrt{399}}{6} \approx 6 - 3.33 = 2.67\) (inside domain)

Verify the maximum

Using the second derivative test:
\[
V''(x) = 24x - 144
\]
At \(x \approx 2.67\):
\[
V''(2.67) = 24(2.67) - 144 \approx -79.92 < 0
\]
Thus, \(x = 6 - \frac{\sqrt{399}}{6}\) yields a local maximum.

Calculate maximum volume

Substitute \(x = 6 - \frac{\sqrt{399}}{6} \approx 2.671\) back into \(V(x)\):
\[
l \approx 23 - 2(2.671) = 17.658\text{ ft}
\]
\[
w \approx 13 - 2(2.671) = 7.658\text{ ft}
\]
\[
V \approx 17.658 \times 7.658 \times 2.671 \approx 361.2\text{ ft}^3
\]
</reasoning>

<answer>
The volume of the largest box that can be formed is approximately \(361.2\text{ ft}^3\) (occurring when \(x = 6 - \frac{\sqrt{399}}{6}\text{ ft} \approx 2.67\text{ ft}\)).
</answer>

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},
0,
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Answer:

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<reasoning>

Express the volume function

The dimensions of the cardboard are \(23\text{ ft}\) by \(13\text{ ft}\).
Cutting out squares of side length \(x\) from each corner yields:

• Length: \(l = 23 - 2x\)
• Width: \(w = 13 - 2x\)
• Height: \(h = x\)

The volume \(V(x)\) is:
\[
V(x) = x(23 - 2x)(13 - 2x) = 4x^3 - 72x^2 + 299x
\]
The physical domain is \(0 < x < 6.5\).

Find the derivative

Differentiate \(V(x)\) with respect to \(x\):
\[
V'(x) = 12x^2 - 144x + 299
\]

Solve for critical points

Set \(V'(x) = 0\):
\[
12x^2 - 144x + 299 = 0
\]
Using the quadratic formula:
\[
x = \frac{144 \pm \sqrt{(-144)^2 - 4(12)(299)}}{24}
\]
\[
x = \frac{144 \pm \sqrt{20736 - 14352}}{24} = \frac{144 \pm \sqrt{6384}}{24}
\]
\[
x = \frac{144 \pm 4\sqrt{399}}{24} = 6 \pm \frac{\sqrt{399}}{6}
\]
Evaluating the roots:

• \(x_1 = 6 + \frac{\sqrt{399}}{6} \approx 6 + 3.33 = 9.33\) (outside domain)
• \(x_2 = 6 - \frac{\sqrt{399}}{6} \approx 6 - 3.33 = 2.67\) (inside domain)

Verify the maximum

Using the second derivative test:
\[
V''(x) = 24x - 144
\]
At \(x \approx 2.67\):
\[
V''(2.67) = 24(2.67) - 144 \approx -79.92 < 0
\]
Thus, \(x = 6 - \frac{\sqrt{399}}{6}\) yields a local maximum.

Calculate maximum volume

Substitute \(x = 6 - \frac{\sqrt{399}}{6} \approx 2.671\) back into \(V(x)\):
\[
l \approx 23 - 2(2.671) = 17.658\text{ ft}
\]
\[
w \approx 13 - 2(2.671) = 7.658\text{ ft}
\]
\[
V \approx 17.658 \times 7.658 \times 2.671 \approx 361.2\text{ ft}^3
\]
</reasoning>

<answer>
The volume of the largest box that can be formed is approximately \(361.2\text{ ft}^3\) (occurring when \(x = 6 - \frac{\sqrt{399}}{6}\text{ ft} \approx 2.67\text{ ft}\)).
</answer>

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