find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard…

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: [ \begin{align*} V(x)&=x(350-50x - 28x+4x^{2})\ &=x(350 - 78x+4x^{2})\ &=4x^{3}-78x^{2}+350x \end{align*} ]

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). [ \begin{align*} x&=\frac{78\pm\sqrt{(-78)^{2}-4\times6\times175}}{2\times6}\ &=\frac{78\pm\sqrt{6084-4200}}{12}\ &=\frac{78\pm\sqrt{1884}}{12}\ &=\frac{78\pm43.4166}{12} \end{align*} ] 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) [ \begin{align*} V&=2.88\times(25-2\times2.88)\times(14 - 2\times2.88)\ &=2.88\times19.24\times8.24\ &\approx459.07 \end{align*} ]

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.