a piece of cardboard measuring 14 inches by 10 inches is formed into an open - top box by cutting squares…

Explanation:

Step1: Determine box dimensions

After cutting $x$ - inch squares from each corner and folding up the sides, the length of the box is $l = 14 - 2x$, the width is $w=10 - 2x$, and the height is $h = x$.

Step2: Find volume formula

The volume formula for a rectangular - box is $V=l\times w\times h$. Substituting the values of $l$, $w$, and $h$ we get $V(x)=(14 - 2x)(10 - 2x)x$. Expand it: [ \begin{align*} V(x)&=(140-28x - 20x+4x^{2})x\ &=(140 - 48x+4x^{2})x\ &=140x-48x^{2}+4x^{3} \end{align*} ]

Step3: Differentiate the volume function

Differentiate $V(x)=4x^{3}-48x^{2}+140x$ with respect to $x$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $V^\prime(x)=12x^{2}-96x + 140$.

Step4: Set the derivative equal to zero

Set $V^\prime(x)=0$, so $12x^{2}-96x + 140 = 0$. Divide through by 4 to simplify: $3x^{2}-24x + 35=0$. Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for a quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 3$, $b=-24$, and $c = 35$. [ \begin{align*} x&=\frac{24\pm\sqrt{(-24)^{2}-4\times3\times35}}{2\times3}\ &=\frac{24\pm\sqrt{576-420}}{6}\ &=\frac{24\pm\sqrt{156}}{6}\ &=\frac{24\pm2\sqrt{39}}{6}\ &=\frac{12\pm\sqrt{39}}{3} \end{align*} ] We also have the physical constraints $14-2x>0$ and $10 - 2x>0$ and $x>0$, which gives $0<x<5$. $x=\frac{12+\sqrt{39}}{3}\approx\frac{12 + 6.245}{3}=\frac{18.245}{3}\approx6.08$ (rejected as it is out of the domain $0<x<5$) $x=\frac{12-\sqrt{39}}{3}\approx\frac{12 - 6.245}{3}=\frac{5.755}{3}\approx1.92$

Answer:

$V(x)=4x^{3}-48x^{2}+140x$ $x=\frac{12 - \sqrt{39}}{3}$