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

Explanation:

Step1: Find surface - area formulas

The surface - area of the hemispherical dome is $A_{dome}=2\pi r^{2}$. Let the height of the cylindrical part be $h$. The surface - area of the cylindrical part (lateral surface) is $A_{cylinder}=2\pi rh$. Let the cost per unit area of concrete be $k$, then the cost per unit area of metal is $2.2k$. The cost function $C$ is $C = 2.2k\times2\pi r^{2}+k\times2\pi rh$.

Step2: Express $h$ in terms of $r$ using volume formula

The volume of the silo $V$ is the sum of the volume of the hemisphere and the volume of the cylinder. $V=\frac{2}{3}\pi r^{3}+\pi r^{2}h$. Given $V = 950$, we have $\pi r^{2}h=950-\frac{2}{3}\pi r^{3}$, so $h=\frac{950-\frac{2}{3}\pi r^{3}}{\pi r^{2}}=\frac{950}{\pi r^{2}}-\frac{2}{3}r$.

Step3: Substitute $h$ into the cost function

Substitute $h=\frac{950}{\pi r^{2}}-\frac{2}{3}r$ into $C = 4.4k\pi r^{2}+2k\pi rh$. [ \begin{align*} C&=4.4k\pi r^{2}+2k\pi r\left(\frac{950}{\pi r^{2}}-\frac{2}{3}r\right)\ &=4.4k\pi r^{2}+\frac{1900k}{r}-\frac{4}{3}k\pi r^{2}\ &=k\left(\left(4.4\pi-\frac{4}{3}\pi\right)r^{2}+\frac{1900}{r}\right)\ &=k\left(\left(\frac{13.2\pi - 4\pi}{3}\right)r^{2}+\frac{1900}{r}\right)\ &=k\left(\frac{9.2\pi}{3}r^{2}+\frac{1900}{r}\right) \end{align*} ] Since $k$ is a non - zero constant (it will not affect the minimization process), we can ignore it and the cost function in terms of $r$ is $C=\frac{9.2\pi}{3}r^{2}+\frac{1900}{r}$.

Answer:

$\frac{9.2\pi}{3}r^{2}+\frac{1900}{r}$