Question

question 18 - 1 point an open - top box is to be made from a 12 - inch by 54 - inch piece of plastic by removing a square from each corner of the plastic and folding up the flaps on each side. what is the maximum possible volume, in cubic inches, of the box? you may enter an exact answer or round to the nearest hundredth. provide your answer below. maximum volume = in³

Explanation:

Step1: Define the variables

Let the side - length of the square removed from each corner be $x$ inches. Then the length of the box is $l = 54 - 2x$ inches, the width of the box is $w=12 - 2x$ inches, and the height of the box is $h = x$ inches. The volume $V$ of the box is given by the function $V(x)=(54 - 2x)(12 - 2x)x$.

Step2: Expand the function

\[

$$\begin{align*} V(x)&=(54\times12-54\times2x-2x\times12 + 4x^{2})x\\ &=(648-108x - 24x+4x^{2})x\\ &=(648 - 132x+4x^{2})x\\ &=648x-132x^{2}+4x^{3} \end{align*}$$

\]

Step3: Find the derivative

Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $V^\prime(x)=648-264x + 12x^{2}$.

Step4: Set the derivative equal to zero

Set $V^\prime(x)=0$, so $12x^{2}-264x + 648 = 0$. Divide through by 12: $x^{2}-22x + 54=0$.

Step5: Solve the quadratic equation

Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b=-22$, and $c = 54$.
\[

$$\begin{align*} x&=\frac{22\pm\sqrt{(-22)^{2}-4\times1\times54}}{2\times1}\\ &=\frac{22\pm\sqrt{484 - 216}}{2}\\ &=\frac{22\pm\sqrt{268}}{2}\\ &=\frac{22\pm2\sqrt{67}}{2}\\ &=11\pm\sqrt{67} \end{align*}$$

\]
We also have the constraints $0\lt x\lt6$ (because if $x = 6$, then $12-2x=0$). Since $11+\sqrt{67}\approx11 + 8.19 = 19.19>6$ and $11-\sqrt{67}\approx11 - 8.19 = 2.81$ which is in the domain $0\lt x\lt6$.

Step6: Find the maximum volume

Substitute $x = 11-\sqrt{67}\approx2.81$ into the volume function $V(x)$.
\[

$$\begin{align*} V(11 - \sqrt{67})&=(54-2(11 - \sqrt{67}))(12-2(11 - \sqrt{67}))(11 - \sqrt{67})\\ &=(54-22 + 2\sqrt{67})(12-22 + 2\sqrt{67})(11 - \sqrt{67})\\ &=(32 + 2\sqrt{67})(-10+2\sqrt{67})(11 - \sqrt{67})\\ \end{align*}$$

\]
\[

$$\begin{align*} V(11 - \sqrt{67})&\approx(32+2\times8.19)(-10 + 2\times8.19)(11-8.19)\\ &=(32 + 16.38)(-10+16.38)(2.81)\\ &=(48.38)(6.38)(2.81)\\ &\approx48.38\times17.93\\ &\approx867.45 \end{align*}$$

\]

Answer:

$867.45$