Answer:
![1.91\ \text{in}](https://img.qammunity.org/2021/formulas/mathematics/college/8kfbz8opj9z8nohqim3f6hfs7cyfkmaz4m.png)
![3.84\ \text{in}](https://img.qammunity.org/2021/formulas/mathematics/college/pyphz9y3deqroo1mrk92pqupvm19124yqk.png)
Explanation:
V = Volume of cylinder =
![44\ \text{in}^3](https://img.qammunity.org/2021/formulas/mathematics/college/6rv051zq7rqn1urzwqzbb5hqwfsl55wvm7.png)
h = Height of cylinder
r = Radius of cylinder
Volume of cylinder is given by
![V=\pi r^2h\\\Rightarrow h=(V)/(\pi r^2)\\\Rightarrow h=(44)/(\pi r^2)](https://img.qammunity.org/2021/formulas/mathematics/college/9y4zv0oa4rhyoea453caslo2xwpdg4k5kl.png)
Total surface area of a cylinder is given by
![S=2\pi r^2+2\pi rh\\\Rightarrow S=2\pi r^2+2\pi r*(44)/(\pi r^2)\\\Rightarrow S=2\pi r^2+(88)/(r)](https://img.qammunity.org/2021/formulas/mathematics/college/wzh2srevx6x2szonp27xgmuurekysczql5.png)
Differentiating with respect to radius
![(dS)/(dr)=4\pi r-(88)/(r^2)](https://img.qammunity.org/2021/formulas/mathematics/college/m1qhbga9xk2iwy99eg6n22tj0kcy067ji6.png)
Equating with zero
![4\pi r-(88)/(r^2)=0\\\Rightarrow 4\pi r=(88)/(r^2)\\\Rightarrow r^3=(88)/(4\pi)\\\Rightarrow r=((22)/(\pi))^{(1)/(3)}\\\Rightarrow r=1.91\ \text{in}](https://img.qammunity.org/2021/formulas/mathematics/college/mjrs3lv3y8m3pumugd0vvrjjidspshqc9v.png)
Double derivative of S
![(d^2S)/(dr^2)=4\pi+176>0](https://img.qammunity.org/2021/formulas/mathematics/college/pqa7y5swaazveoswaagi7uuanlhaae58dz.png)
So
is minimum at
![(dS)/(dr)=0](https://img.qammunity.org/2021/formulas/mathematics/college/ee56o8ctpz7svotgf3r9sxk7daz4ycjgi2.png)
![h=(44)/(\pi r^2)=(44)/(\pi 1.91^2)\\\Rightarrow h=3.84\ \text{in}](https://img.qammunity.org/2021/formulas/mathematics/college/czjh56agyht80k5aseco9bujwbx01z7zud.png)
So the radius and height of the cylinder is
and
respectively such that the least amount of metal is used.