a supermarket employee wants to construct an open - top box from a 10 by 16 in piece of cardboard. to do…

Explanation:

Step1: Define the variables

Let the side - length of the square cut from each corner be $x$ inches. The length of the box will be $l = 16 - 2x$ inches, the width will be $w=10 - 2x$ inches, and the height will be $h = x$ inches. The volume $V$ of the box is given by the function $V(x)=(16 - 2x)(10 - 2x)x$.

Step2: Expand the function

[ \begin{align*} V(x)&=(16 - 2x)(10 - 2x)x\ &=(160-32x - 20x+4x^{2})x\ &=(160 - 52x+4x^{2})x\ &=160x-52x^{2}+4x^{3} \end{align*} ]

Step3: Find the derivative

Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $V^\prime(x)=160-104x + 12x^{2}$.

Step4: Set the derivative equal to zero

$12x^{2}-104x + 160=0$. Divide through by 4: $3x^{2}-26x + 40=0$.

Step5: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 3$, $b=-26$, $c = 40$), the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. [ \begin{align*} x&=\frac{26\pm\sqrt{(-26)^{2}-4\times3\times40}}{2\times3}\ &=\frac{26\pm\sqrt{676 - 480}}{6}\ &=\frac{26\pm\sqrt{196}}{6}\ &=\frac{26\pm14}{6} \end{align*} ] We get two solutions: $x_1=\frac{26 + 14}{6}=\frac{40}{6}=\frac{20}{3}\approx6.67$ and $x_2=\frac{26-14}{6}=2$. But $x=\frac{20}{3}$ is not valid since if $x=\frac{20}{3}$, then $10-2x=10-\frac{40}{3}=-\frac{10}{3}<0$ and $16 - 2x=16-\frac{40}{3}=\frac{8}{3}$.

Answer:

A. 2 in