a manufacturer cuts squares from the corners of a rectangular piece of sheet metal that measures 3 inches by…

Explanation:

Step1: Find the volume formula

The length of the box after cutting is $l = 7 - 2x$, the width is $w=3 - 2x$ and the height is $h = x$. The volume $V(x)=(7 - 2x)(3 - 2x)x=4x^{3}-20x^{2}+21x$, where $0<x<1.5$ (since $3 - 2x>0$).

Step2: Differentiate the volume function

Differentiate $V(x)$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $V'(x)=12x^{2}-40x + 21$.

Step3: Find the critical points

Set $V'(x)=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 12$, $b=-40$, $c = 21$. So $x=\frac{40\pm\sqrt{(-40)^{2}-4\times12\times21}}{2\times12}=\frac{40\pm\sqrt{1600 - 1008}}{24}=\frac{40\pm\sqrt{592}}{24}=\frac{40\pm4\sqrt{37}}{24}=\frac{10\pm\sqrt{37}}{6}$. We have $x_1=\frac{10+\sqrt{37}}{6}\approx\frac{10 + 6.083}{6}\approx2.68$ (rejected as it is out of the domain $0<x<1.5$) and $x_2=\frac{10-\sqrt{37}}{6}\approx\frac{10 - 6.083}{6}\approx0.65$.

Step4: Find the maximum volume

Substitute $x\approx0.65$ into the volume function $V(x)$. $V(0.65)=4\times(0.65)^{3}-20\times(0.65)^{2}+21\times0.65=4\times0.274625-20\times0.4225+13.65=1.0985 - 8.45+13.65 = 6.2985\approx6.30$.

Answer:

Value of $x$ that maximizes volume: $0.65$ in Maximum volume: $6.30$ in$^{3}$