find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard…

Explanation:

Step1: Let side - length of square

Let the side - length of the square cut from each corner be $x$ inches. Then the length of the box $L=(19 - 2x)$ inches, the width of the box $W=(11 - 2x)$ inches, and the height of the box $H = x$ inches. The volume $V$ of the box is given by the function $V(x)=(19 - 2x)(11 - 2x)x$. [ \begin{align*} V(x)&=(19\times11-19\times2x-2x\times11 + 4x^{2})x\ &=(209-38x - 22x+4x^{2})x\ &=(209x-60x^{2}+4x^{3}) \end{align*} ]

Step2: Find the derivative

Differentiate $V(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$. [ V^\prime(x)=209-120x + 12x^{2} ]

Step3: Set the derivative equal to zero

To find the critical points, set $V^\prime(x)=0$, so $12x^{2}-120x + 209 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, where $a = 12$, $b=-120$, and $c = 209$. [ x=\frac{120\pm\sqrt{(-120)^{2}-4\times12\times209}}{2\times12}=\frac{120\pm\sqrt{14400-10032}}{24}=\frac{120\pm\sqrt{4368}}{24}=\frac{120\pm66.09}{24} ] We get $x_1=\frac{120 + 66.09}{24}\approx7.75$ and $x_2=\frac{120 - 66.09}{24}\approx2.25$. But $x = 7.75$ is not valid since $11-2x=11 - 2\times7.75=11 - 15.5=-4.5<0$. So we take $x = 2.25$.

Step4: Find the dimensions

Length $L=19-2x=19-2\times2.25 = 19 - 4.5=14.5$ inches, width $W=11-2x=11 - 2\times2.25=11 - 4.5 = 6.5$ inches, and height $H=x = 2.25$ inches.

Step5: Find the volume

Substitute $x = 2.25$ into the volume formula $V=(19 - 2x)(11 - 2x)x$. [ V=(14.5)\times(6.5)\times(2.25)=14.5\times14.625 = 211.0625\approx211.31\text{ in}^3 ]

Answer:

The dimensions of box of maximum volume are $14.5,6.5,2.25$ in. The maximum volume is $211.31$ in$^{3}$