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 with length $l = 3x$, width $w=x$, and height $h$ is $V=lwh$. Given $V = 12$, so $12=3x\cdot x\cdot h=3x^{2}h$. Then we can express $h=\frac{12}{3x^{2}}=\frac{4}{x^{2}}$.

Step2: Find the surface - area formulas

The area of the top and bottom is $A_{tb}=2\times(3x\cdot x)=6x^{2}$. The area of the four sides is $A_{s}=2(3x\cdot h)+2(x\cdot h)=2h(3x + x)=8xh$.

Step3: Find the cost function

The cost function $C(x)$ is the sum of the cost of the top and bottom and the cost of the four sides. The cost of the top and bottom is $2.5\times6x^{2}=15x^{2}$, and the cost of the four sides is $2.7\times8xh = 21.6xh$. Substitute $h=\frac{4}{x^{2}}$ into the cost function: $C(x)=15x^{2}+21.6x\cdot\frac{4}{x^{2}}=15x^{2}+\frac{86.4}{x},x>0$.

Step4: Find the derivative of the cost function

Differentiate $C(x)$ with respect to $x$. Using the power rule, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. So $C^\prime(x)=30x-\frac{86.4}{x^{2}}$.

Step5: Set the derivative equal to zero and solve for $x$

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 $x=\sqrt[3]{2.88}\approx1.42$.

Step6: Find the height $h$

Substitute $x\approx1.42$ into the formula for $h$. Since $h=\frac{4}{x^{2}}$, then $h=\frac{4}{(1.42)^{2}}=\frac{4}{2.0164}\approx1.98$.

Answer:

$1.98$ cm