Question

1. if 12 ft² of material is available to make a box with square base and open top, find the largest possible volume for the box. (make sure to justify why your answer corresponds to an absolute maximum.)

Explanation:

Step1: Set up the surface - area and volume equations

Let the side length of the square base be \(x\) (in feet) and the height of the box be \(y\) (in feet).
The surface - area of the open - top box with a square base is \(S = x^{2}+4xy\). Given \(S = 12\), so \(12=x^{2}+4xy\), and we can express \(y=\frac{12 - x^{2}}{4x}\).
The volume of the box is \(V=x^{2}y\). Substitute \(y\) into the volume formula: \(V(x)=x^{2}\cdot\frac{12 - x^{2}}{4x}=\frac{12x - x^{3}}{4}=3x-\frac{1}{4}x^{3}\), where \(x>0\).

Step2: Find the first - derivative of the volume function

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

Step3: Find the critical points

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

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Step4: Find the second - derivative of the volume function

Differentiate \(V^\prime(x)\) with respect to \(x\). \(V^{\prime\prime}(x)=-\frac{3}{2}x\).

Step5: Use the second - derivative test

When \(x = 2\), \(V^{\prime\prime}(2)=-\frac{3}{2}\times2=-3<0\). So the function \(V(x)\) has a local maximum at \(x = 2\).

Step6: Find the maximum volume

Substitute \(x = 2\) into the volume formula \(V(x)\).
When \(x = 2\), \(y=\frac{12-2^{2}}{4\times2}=\frac{12 - 4}{8}=1\).
\(V(2)=2^{2}\times1 = 4\) (in cubic feet).

Answer:

The largest possible volume of the box is \(4\space ft^{3}\).