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: Define variables

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 formula for the open - top box 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 formula for the box is \(V=x^{2}y\).

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

Substitute \(y=\frac{12 - x^{2}}{4x}\) into \(V=x^{2}y\). Then \(V(x)=x^{2}\cdot\frac{12 - x^{2}}{4x}=\frac{12x - x^{3}}{4}=3x-\frac{1}{4}x^{3}\), where \(x>0\).

Step3: Find the first derivative

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

Step4: Find the critical points

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

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Step5: Find the second derivative

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

Step6: Find the value of \(y\) and \(V\)

When \(x = 2\), \(y=\frac{12-2^{2}}{4\times2}=\frac{12 - 4}{8}=1\).
The volume \(V=x^{2}y=(2)^{2}\times1 = 4\) (in \(ft^{3}\)).

Step7: Justify the absolute maximum

Since \(x>0\), and as \(x
ightarrow0^{+}\), \(V(x)=3x-\frac{1}{4}x^{3}
ightarrow0\); as \(x
ightarrow\sqrt{12}\) (when \(y = 0\)), \(V(x)=3x-\frac{1}{4}x^{3}
ightarrow0\). And we have found a local maximum at \(x = 2\) (using the second - derivative test \(V^{\prime\prime}(2)<0\)), so this local maximum is also an absolute maximum.

Answer:

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