you are making a rectangular box out of a 12 - inch by 8 - inch piece of cardboard. the box will be formed…

Explanation:

Step1: Find the volume function

The length of the box after cutting and folding is $l = 12 - 2x$, the width is $w=8 - 2x$, and the height is $h = x$. The volume $V$ of the rectangular - box is given by the function $V(x)=(12 - 2x)(8 - 2x)x=4x^{3}-40x^{2}+96x$, where $0\lt x\lt4$ (because $12-2x\gt0$ and $8 - 2x\gt0$).

Step2: Find the derivative of the volume function

Using the power - rule $(x^{n})^\prime=nx^{n - 1}$, we have $V^\prime(x)=12x^{2}-80x + 96$.

Step3: Set the derivative equal to zero and solve for x

$12x^{2}-80x + 96 = 0$. Divide through by 4: $3x^{2}-20x + 24 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 3$, $b=-20$, and $c = 24$. So $x=\frac{20\pm\sqrt{(-20)^{2}-4\times3\times24}}{2\times3}=\frac{20\pm\sqrt{400 - 288}}{6}=\frac{20\pm\sqrt{112}}{6}=\frac{20\pm4\sqrt{7}}{6}=\frac{10\pm2\sqrt{7}}{3}$. $x_1=\frac{10 + 2\sqrt{7}}{3}\approx\frac{10+2\times2.646}{3}=\frac{10 + 5.292}{3}\approx5.097$ (rejected since $x\lt4$) and $x_2=\frac{10 - 2\sqrt{7}}{3}\approx\frac{10-5.292}{3}\approx1.57$. So $x\approx1.6$ inches.

Step4: Find the maximum volume

Substitute $x\approx1.6$ into the volume function $V(x)$. $V(1.6)=(12-2\times1.6)(8 - 2\times1.6)\times1.6=(12 - 3.2)(8 - 3.2)\times1.6=8.8\times4.8\times1.6=8.8\times7.68 = 67.584\approx67.6$ $in^{3}$.

Step5: Find the dimensions of the box

Length $l=12-2x\approx12-2\times1.6 = 12 - 3.2 = 8.8$ inches. Width $w=8-2x\approx8 - 2\times1.6=8 - 3.2 = 4.8$ inches. Height $h=x\approx1.6$ inches.

Answer:

a. $1.6$ b. $67.6$ c. length is about $8.8$ in, width is about $4.8$ in, height is about $1.6$ in