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 = 9534.13 × 3510.5660617732 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 \(S=x^{2}+4xh\) (since the top is open), and \(S = 1100\), so \(h=\frac{1100 - x^{2}}{4x}\).
The volume \(V=x^{2}h\).

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

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

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: Set the derivative equal to zero

Set \(V^\prime = 0\), then \(\frac{1}{4}(1100 - 3x^{2})=0\).
Multiply both sides by \(4\) to get \(1100-3x^{2}=0\).
Solve for \(x\): \(3x^{2}=1100\), \(x^{2}=\frac{1100}{3}\), \(x=\sqrt{\frac{1100}{3}}\approx19.15\) (we take the positive value since \(x\) represents a length).

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

Differentiate \(V^\prime=\frac{1}{4}(1100 - 3x^{2})\) 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 the volume is maximized at this \(x\) - value.

Step6: Find the value of \(h\)

Substitute \(x = \sqrt{\frac{1100}{3}}\) into \(h=\frac{1100 - x^{2}}{4x}\).
Since \(x^{2}=\frac{1100}{3}\), then \(h=\frac{1100-\frac{1100}{3}}{4\sqrt{\frac{1100}{3}}}=\frac{\frac{2200}{3}}{4\sqrt{\frac{1100}{3}}}=\sqrt{\frac{1100}{12}}\approx9.58\).

Step7: Calculate the volume

Substitute \(x=\sqrt{\frac{1100}{3}}\) and \(h=\sqrt{\frac{1100}{12}}\) into \(V=x^{2}h\).
\(V=\frac{1100}{3}\times\sqrt{\frac{1100}{12}}=\frac{1100}{3}\times\frac{\sqrt{1100}}{\sqrt{12}}=\frac{1100}{3}\times\frac{10\sqrt{11}}{\ 2\sqrt{3}}=\frac{5500\sqrt{33}}{9}\approx3510.57\)

Answer:

\(3510.57\) cubic centimeters.