Question

a covered box is to be made from a rectangular sheet of cardboard measuring 30 inches by 48 inches. this is done by cutting out the shaded regions of the figure and then folding on the dotted lines. what are the dimensions x, y, and z that maximize the volume? x = inches y = inches z = inches

Explanation:

Step1: Express \(y\) and \(z\) in terms of \(x\)

From the cardboard dimensions:

• The length \(z = 30 - 2x\) (since we cut \(x\) from both ends of the 30 - inch side)
• The width \(y=48 - 2x\) (since we cut \(x\) from both ends of the 48 - inch side)

Step2: Write the volume formula

The volume \(V\) of a rectangular box is \(V=x\times y\times z\). Substituting \(y = 48 - 2x\) and \(z = 30 - 2x\) into the volume formula, we get \(V=x(48 - 2x)(30 - 2x)\)

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Step3: Find the derivative of \(V\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(V^\prime\) is:

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Step4: Set \(V^\prime = 0\) and solve for \(x\)

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Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-26\), \(c = 120\))

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We get \(x_1=\frac{26 + 14}{2}=20\) and \(x_2=\frac{26-14}{2}=6\)

Step5: Use the second - derivative test

Find the second - derivative \(V^{\prime\prime}=(12x^{2}-312x + 1440)^\prime=24x-312\)

• When \(x = 20\), \(V^{\prime\prime}(20)=24\times20-312=480 - 312=168>0\) (function has a minimum at \(x = 20\))
• When \(x = 6\), \(V^{\prime\prime}(6)=24\times6-312=144-312=-168<0\) (function has a maximum at \(x = 6\))

Step6: Calculate \(y\) and \(z\)

• When \(x = 6\)
• \(y=48-2x=48-2\times6 = 36\)
• \(z=30-2x=30 - 2\times6=18\)

Answer:

\(x = 6\) inches, \(y = 36\) inches, \(z = 18\) inches