Question

question 2
if 2300 square centimeters of material is available to make a box with a square base and an open top, find
the largest possible volume of the box.
note: this is similar to problem 1, only the material (surface area) is the constraint, rather than the
volume. the volume is now the objective function - the quantity we wish to maximize.
volume =
cubic centimeters.
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Explanation:

Step1: Express surface area and volume

Let the side length of the square base be \(x\) cm and the height of the box be \(h\) cm.
The surface area \(S=x^{2}+4xh\) (since the box has an open - top), and \(S = 2300\), so \(h=\frac{2300 - x^{2}}{4x}\).
The volume \(V=x^{2}h\). Substitute \(h\) into the volume formula: \(V=x^{2}\cdot\frac{2300 - x^{2}}{4x}=\frac{2300x - x^{3}}{4}=\frac{2300}{4}x-\frac{1}{4}x^{3}\).

Step2: Find the derivative of the volume function

Differentiate \(V(x)\) with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(V^\prime(x)=\frac{2300}{4}-\frac{3}{4}x^{2}\).

Step3: Set the derivative equal to zero and solve for \(x\)

Set \(V^\prime(x) = 0\), then \(\frac{2300}{4}-\frac{3}{4}x^{2}=0\).
Multiply through by \(4\) to get \(2300-3x^{2}=0\).
Rearrange: \(3x^{2}=2300\), so \(x^{2}=\frac{2300}{3}\), and \(x=\sqrt{\frac{2300}{3}}\) (we take the positive value since \(x\) represents a length).

Step4: Check the second - derivative

Differentiate \(V^\prime(x)\) to get \(V^{\prime\prime}(x)=-\frac{3}{2}x\). When \(x = \sqrt{\frac{2300}{3}}\), \(V^{\prime\prime}(x)<0\), which means \(V(x)\) has a maximum at this \(x\) - value.

Step5: Calculate the maximum volume

Substitute \(x=\sqrt{\frac{2300}{3}}\) into the volume formula.
\(V=\frac{2300}{4}\sqrt{\frac{2300}{3}}-\frac{1}{4}(\sqrt{\frac{2300}{3}})^{3}\)
\(V=\frac{1}{4}\sqrt{\frac{2300^{3}}{3}}-\frac{1}{4}\sqrt{\frac{2300^{3}}{3^{3}}}\)
\(V=\frac{2}{4}\sqrt{\frac{2300^{3}}{3^{3}}}\)
Another way:
Since \(h=\frac{2300 - x^{2}}{4x}\) and \(x^{2}=\frac{2300}{3}\), then \(h=\frac{2300-\frac{2300}{3}}{4\sqrt{\frac{2300}{3}}}=\frac{\frac{4600}{3}}{4\sqrt{\frac{2300}{3}}}=\sqrt{\frac{2300}{3}}\)
\(V=x^{2}h\), and since \(x^{2}=\frac{2300}{3}\) and \(h = \sqrt{\frac{2300}{3}}\)
\(V=\frac{2300}{3}\sqrt{\frac{2300}{3}}=\frac{2300\sqrt{6900}}{9}\approx\frac{2300\times83.066}{9}\approx21253.9\)

Answer:

\(\frac{2300\sqrt{6900}}{9}\approx21253.9\) cubic centimeters.