Question

a piece of cardboard measuring 10 inches by 12 inches is formed into an open - top box by cutting squares with side length x from each corner and folding up the sides. find a formula for the volume of the box in terms of x. v(x)= find the value for x that will maximize the volume of the box (round your answer to the nearest hundredths place.) x = inches question help: d post to forum submit question jump to answer

Explanation:

Step1: Determine box dimensions

After cutting $x$ from each corner, the length of the box is $l = 12 - 2x$, 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 $V(x)=(12 - 2x)(10 - 2x)x$.
Expanding, we have:
\[

$$\begin{align*} V(x)&=(120-24x - 20x+4x^{2})x\\ &=(120 - 44x + 4x^{2})x\\ &=120x-44x^{2}+4x^{3} \end{align*}$$

\]

Step3: Differentiate the volume function

Differentiate $V(x)=4x^{3}-44x^{2}+120x$ with respect to $x$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we get $V^\prime(x)=12x^{2}-88x + 120$.

Step4: Set the derivative equal to zero

Set $V^\prime(x)=0$, so $12x^{2}-88x + 120 = 0$. Divide through by 4 to simplify: $3x^{2}-22x + 30=0$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 3$, $b=-22$, and $c = 30$.
\[

$$\begin{align*} x&=\frac{22\pm\sqrt{(-22)^{2}-4\times3\times30}}{2\times3}\\ &=\frac{22\pm\sqrt{484 - 360}}{6}\\ &=\frac{22\pm\sqrt{124}}{6}\\ &=\frac{22\pm2\sqrt{31}}{6}\\ &=\frac{11\pm\sqrt{31}}{3} \end{align*}$$

\]
$x_1=\frac{11+\sqrt{31}}{3}\approx\frac{11 + 5.57}{3}=\frac{16.57}{3}\approx5.52$ and $x_2=\frac{11-\sqrt{31}}{3}\approx\frac{11 - 5.57}{3}=\frac{5.43}{3}\approx1.81$.
But $x$ must satisfy $0\lt x\lt5$ (because if $x = 5$, then $10-2x=0$). So we reject $x\approx5.52$.

Answer:

$V(x)=4x^{3}-44x^{2}+120x$
$x\approx1.81$