a grain silo consists of a cylindrical concrete tower surrounded by a metal hemispherical dome. the metal in…

Explanation:

Step1: Recall volume formulas

The volume of the silo $V = \pi r^{2}h+\frac{2}{3}\pi r^{3}$, and $V = 950$. So $\pi r^{2}h+\frac{2}{3}\pi r^{3}=950$, then $h=\frac{950 - \frac{2}{3}\pi r^{3}}{\pi r^{2}}$.

Step2: Recall cost - surface area relationship

The cost function $C$ (assuming cost per unit area of concrete is 1) for the surface area of the silo: The surface area of the cylindrical part is $2\pi rh$ and the surface area of the hemispherical part is $2\pi r^{2}$, and since the metal in the dome costs 2.2 times as much as the concrete per unit surface - area, $C = 2\pi rh+2.2\times2\pi r^{2}$. Substitute $h=\frac{950 - \frac{2}{3}\pi r^{3}}{\pi r^{2}}$ into the cost function: [ \begin{align*} C&=2\pi r\times\frac{950 - \frac{2}{3}\pi r^{3}}{\pi r^{2}}+4.4\pi r^{2}\ &=\frac{1900}{r}-\frac{4}{3}\pi r^{2}+4.4\pi r^{2}\ &=\frac{1900}{r}+\left(4.4\pi-\frac{4}{3}\pi\right)r^{2}\ &=\frac{1900}{r}+\frac{13.2\pi - 4\pi}{3}r^{2}\ &=\frac{1900}{r}+\frac{9.2\pi}{3}r^{2} \end{align*} ] We are given that the radius $r = 4.6$ m.

Step3: Calculate the height

Substitute $r = 4.6$ into the volume - height formula $h=\frac{950 - \frac{2}{3}\pi r^{3}}{\pi r^{2}}$. [ \begin{align*} h&=\frac{950-\frac{2}{3}\pi\times(4.6)^{3}}{\pi\times(4.6)^{2}}\ &=\frac{950-\frac{2}{3}\pi\times97.336}{\pi\times21.16}\ &=\frac{950 - 203.97}{\ 66.48}\ &=\frac{746.03}{66.48}\ &\approx11.2 \end{align*} ]

Answer:

$11.2$ m