Question

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if 1100 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 = 9535.94 × cubic centimeters.
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Explanation:

Step1: Define variables

Let the side length of the square base be \(x\) cm and the height of the box be \(y\) cm.
The surface area \(S=x^{2}+4xy\) (since it has a square base and an open - top), and \(S = 1100\), so \(y=\frac{1100 - x^{2}}{4x}\).
The volume \(V=x^{2}y\).

Step2: Substitute \(y\) into the volume formula

Substitute \(y=\frac{1100 - x^{2}}{4x}\) into \(V=x^{2}y\), we get \(V=\frac{1}{4}(1100x - x^{3})\).

Step3: Find the derivative of \(V\)

Differentiate \(V\) with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(V^\prime=\frac{1}{4}(1100 - 3x^{2})\).

Step4: Find the critical points

Set \(V^\prime = 0\), then \(\frac{1}{4}(1100 - 3x^{2})=0\).

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We discard the negative root since \(x>0\).

Step5: Check the second - derivative

Differentiate \(V^\prime\) with respect to \(x\), \(V^{\prime\prime}=\frac{1}{4}(-6x)=-\frac{3}{2}x\). When \(x = \sqrt{\frac{1100}{3}}\), \(V^{\prime\prime}<0\), so \(V\) has a maximum at this \(x\).

Step6: Find the value of \(y\)

Substitute \(x=\sqrt{\frac{1100}{3}}\) into \(y=\frac{1100 - x^{2}}{4x}\).

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Step7: Calculate the volume

Substitute \(x=\sqrt{\frac{1100}{3}}\) and \(y = \frac{\sqrt{1100}}{6}\) into \(V=x^{2}y\).

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Answer:

\(9534.13\) cubic centimeters.