a container manufacturer plans to make rectangular boxes whose bottom and top measure 3x by x. the container…

Explanation:

Step1: Find the volume formula

The volume $V$ of a rectangular - box is $V = l\times w\times h$. Given $l = 3x$, $w = x$, and $V=12$, so $12 = 3x\cdot x\cdot h=3x^{2}h$. Then $h=\frac{4}{x^{2}}$.

Step2: Calculate the surface - area of the top and bottom

The area of the top and bottom $A_{tb}=2\times(3x\cdot x)=6x^{2}$.

Step3: Calculate the surface - area of the sides

The area of the four sides $A_{s}=2(3x\cdot h + x\cdot h)=2h(3x + x)=8xh$. Substitute $h = \frac{4}{x^{2}}$ into it, we get $A_{s}=8x\cdot\frac{4}{x^{2}}=\frac{32}{x}$.

Step4: Set up the cost function

The cost function $C$ is $C = 2.5A_{tb}+2.7A_{s}$. Substitute $A_{tb}=6x^{2}$ and $A_{s}=\frac{32}{x}$ into it, we have $C = 2.5\times6x^{2}+2.7\times\frac{32}{x}=15x^{2}+\frac{86.4}{x}$.

Step5: Find the derivative of the cost function

Differentiate $C(x)$ with respect to $x$. $C^\prime(x)=30x-\frac{86.4}{x^{2}}$.

Step6: Set the derivative equal to zero

Set $C^\prime(x) = 0$, then $30x-\frac{86.4}{x^{2}} = 0$. Multiply through by $x^{2}$ to get $30x^{3}-86.4 = 0$. Then $x^{3}=\frac{86.4}{30}=2.88$, and $x=\sqrt[3]{2.88}\approx1.42$.

Step7: Find the height

Substitute $x\approx1.42$ into $h=\frac{4}{x^{2}}$. $h=\frac{4}{(1.42)^{2}}=\frac{4}{2.0164}\approx1.98$.

Answer:

$1.98$ cm