Question

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 = 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 \(h\) cm.
The surface area \(S\) of the open - top box is \(S=x^{2}+4xh\). Given \(S = 1100\), so \(1100=x^{2}+4xh\), and \(h=\frac{1100 - x^{2}}{4x}\).
The volume \(V\) of the box is \(V=x^{2}h\).

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

Substitute \(h=\frac{1100 - x^{2}}{4x}\) into \(V=x^{2}h\), we get \(V=x^{2}\times\frac{1100 - x^{2}}{4x}=\frac{1100x - x^{3}}{4}=275x-\frac{1}{4}x^{3}\).

Step3: Find the derivative of \(V\)

Differentiate \(V\) with respect to \(x\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(V^\prime=275-\frac{3}{4}x^{2}\).

Step4: Set the derivative equal to zero

Set \(V^\prime = 0\), then \(275-\frac{3}{4}x^{2}=0\).

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Step5: Find the second - derivative of \(V\)

Differentiate \(V^\prime=275-\frac{3}{4}x^{2}\) with respect to \(x\), \(V^{\prime\prime}=-\frac{3}{2}x\).
When \(x = \sqrt{\frac{1100}{3}}\), \(V^{\prime\prime}<0\), so \(V\) has a maximum at this \(x\) value.

Step6: Calculate the maximum volume

Substitute \(x=\sqrt{\frac{1100}{3}}\) into \(V = 275x-\frac{1}{4}x^{3}\).

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Answer:

\(9535.94\)