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

Explanation:

Step1: Define the variables

Let the side - length of the square cut from each corner be $x$ inches. The length of the box $L=(21 - 2x)$ inches, the width of the box $W=(12 - 2x)$ inches, and the height of the box $H = x$ inches. The volume $V$ of the box is given by the function $V(x)=(21 - 2x)(12 - 2x)x$.

Step2: Expand the volume function

[ \begin{align*} V(x)&=(21 - 2x)(12 - 2x)x\ &=(252-42x - 24x+4x^{2})x\ &=(252x-66x^{2}+4x^{3}) \end{align*} ]

Step3: Find the derivative of the volume function

Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $V^\prime(x)=252-132x + 12x^{2}$.

Step4: Set the derivative equal to zero and solve for $x$

[ \begin{align*} 12x^{2}-132x + 252&=0\ x^{2}-11x + 21&=0 \end{align*} ] Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, where $a = 1$, $b=-11$, and $c = 21$. [ \begin{align*} x&=\frac{11\pm\sqrt{(-11)^{2}-4\times1\times21}}{2\times1}\ &=\frac{11\pm\sqrt{121 - 84}}{2}\ &=\frac{11\pm\sqrt{37}}{2}\ x_1&=\frac{11+\sqrt{37}}{2}\approx\frac{11 + 6.083}{2}=8.54\ x_2&=\frac{11-\sqrt{37}}{2}\approx\frac{11 - 6.083}{2}=2.46 \end{align*} ] But $x = 8.54$ is not valid since $12-2x=12-2\times8.54=-5.08\lt0$. So we take $x\approx2.46$.

Step5: Find the dimensions of the box

Length $L=21-2x\approx21-2\times2.46 = 21 - 4.92=16.08$ inches. Width $W=12-2x\approx12-2\times2.46=12 - 4.92 = 7.08$ inches. Height $H=x\approx2.46$ inches.

Step6: Find the volume of the box

$V = L\times W\times H\approx16.08\times7.08\times2.46\approx277.64$ cubic - inches.

Answer:

Length: $16.08$ in., Width: $7.08$ in., Height: $2.46$ in., Volume: $277.64$ in.$^{3}$