Question

find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 25 in. by 14 in. by cutting congruent squares from the corners and folding up the sides. then find the volume
the dimensions of box of maximum volume are
(round to the nearest hundredth as needed. use a comma to separate answers as needed.)
the maximum volume is
(round to the nearest hundredth as needed.)

Explanation:

Step1: Define variables

Let the side length of the square cut from each corner be \(x\) inches. Then the length of the box is \(L=(25 - 2x)\) inches, the width is \(W=(14 - 2x)\) inches, and the height is \(H = x\) inches. The volume \(V\) of the box is given by the formula \(V(x)=x(25 - 2x)(14 - 2x)\).
Expand the function:

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Step2: Find the derivative

Differentiate \(V(x)\) with respect to \(x\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(V^\prime(x)=12x^{2}-156x + 350\)

Step3: Set the derivative equal to zero

Set \(V^\prime(x)=0\), so \(12x^{2}-156x + 350=0\). Divide through by \(2\): \(6x^{2}-78x + 175=0\).
Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 6\), \(b=-78\), and \(c = 175\).

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We get \(x_1=\frac{78 + 43.4166}{12}\approx10.12\) and \(x_2=\frac{78-43.4166}{12}\approx2.88\).
But \(x = 10.12\) is not valid since \(14-2x=14-2\times10.12=-6.24\lt0\). So we consider \(x\approx2.88\).

Step4: Find the dimensions

Length \(L=25-2x=25-2\times2.88 = 19.24\) inches, width \(W=14 - 2x=14-2\times2.88=8.24\) inches, height \(H=x = 2.88\) inches.

Step5: Find the volume

\(V=x(25 - 2x)(14 - 2x)\). Substitute \(x = 2.88\)

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Answer:

The dimensions of the box of maximum volume are \(19.24\) in, \(8.24\) in, \(2.88\) in. The maximum volume is \(459.07\) in³.