Question

score on last try: 0 of 4 pts. see details for more.
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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 = 9534.13 × 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 box (open - top) is \(S=x^{2}+4xh\). Given \(S = 1100\), so \(1100=x^{2}+4xh\), and we can express \(h=\frac{1100 - x^{2}}{4x}\).
The volume \(V\) of the box is \(V=x^{2}h\). Substitute \(h\) into the volume formula: \(V=x^{2}\cdot\frac{1100 - x^{2}}{4x}=\frac{1100x - x^{3}}{4}=\frac{1100}{4}x-\frac{1}{4}x^{3}=275x-\frac{1}{4}x^{3}\).

Step2: Find the derivative of the volume function

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

Step3: Find the critical points

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

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We ignore the negative root since \(x>0\) (side length).

Step4: Check the second - derivative

Differentiate \(V^\prime(x)\) to get \(V^{\prime\prime}(x)=-\frac{3}{2}x\). When \(x = \sqrt{\frac{1100}{3}}\), \(V^{\prime\prime}(x)<0\), so the volume function has a maximum at this \(x\) value.

Step5: Calculate the height and volume

Substitute \(x=\sqrt{\frac{1100}{3}}\) into the formula for \(h\): \(h=\frac{1100 - x^{2}}{4x}\). Since \(x^{2}=\frac{1100}{3}\), then \(h=\frac{1100-\frac{1100}{3}}{4x}=\frac{\frac{2200}{3}}{4x}=\frac{550}{3x}\).
Substitute \(x=\sqrt{\frac{1100}{3}}\) into the volume formula \(V = 275x-\frac{1}{4}x^{3}\).

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Answer:

\(9534.13\) cubic centimeters.