Answer:
Given :Cost function :
![C(x)=7.6x + 10,800](https://img.qammunity.org/2020/formulas/mathematics/college/enrbkps0376128rvh0v24rxcnw03ucakm4.png)
To Find : Find the average cost per item when the required number of items is produced
Solution:
a) 200 items
Cost function :
![C(x)=7.6x + 10,800](https://img.qammunity.org/2020/formulas/mathematics/college/enrbkps0376128rvh0v24rxcnw03ucakm4.png)
Substitute x = 200
![C(200)=7.6(200)+ 10,800](https://img.qammunity.org/2020/formulas/mathematics/college/34wjh8l3ixbbs1rnn3qz2zh1j1s2rqksnf.png)
![C(200)=12320](https://img.qammunity.org/2020/formulas/mathematics/college/eqddust8jar02pmfvjzhx6a4gsthat9ot1.png)
So, Total cost of producing 200 items is 12320
Now Average cost per item =
![\frac{\text{Total cost}}{\text{No. of items}}](https://img.qammunity.org/2020/formulas/mathematics/college/u4za6lmg4patn0qd421ldaa9mtucy3p659.png)
Average cost per item=
![(12320)/(200)](https://img.qammunity.org/2020/formulas/mathematics/college/s84mwnfcitmwfsazuvy2wj9ufxzl6h94ah.png)
Average cost per item=61.6
So, the average cost per item when 200 items are produced is 61.6 .
b) 2000 items
Cost function :
![C(x)=7.6x + 10,800](https://img.qammunity.org/2020/formulas/mathematics/college/enrbkps0376128rvh0v24rxcnw03ucakm4.png)
Substitute x = 2000
![C(2000)=7.6(2000)+ 10,800](https://img.qammunity.org/2020/formulas/mathematics/college/pwehpbrd9hypnt847o3g1mo7ozvmpnfy62.png)
![C(2000)=26000](https://img.qammunity.org/2020/formulas/mathematics/college/8nireg49ph98aa86c542enjr38z4zujde7.png)
So, Total cost of producing 2000 items is 26000
Now Average cost per item =
![\frac{\text{Total cost}}{\text{No. of items}}](https://img.qammunity.org/2020/formulas/mathematics/college/u4za6lmg4patn0qd421ldaa9mtucy3p659.png)
Average cost per item=
![(26000)/(2000)](https://img.qammunity.org/2020/formulas/mathematics/college/41go9lq6ufp6pxwemt9kco438o1nc6wyiw.png)
Average cost per item=13
So, the average cost per item when 2000 items are produced is 13.
c) 5000 items
Cost function :
![C(x)=7.6x + 10,800](https://img.qammunity.org/2020/formulas/mathematics/college/enrbkps0376128rvh0v24rxcnw03ucakm4.png)
Substitute x = 5000
![C(5000)=7.6(5000)+ 10,800](https://img.qammunity.org/2020/formulas/mathematics/college/a93iizciua5y0r9kziwig57kgcefsaa0ac.png)
![C(5000)=48800](https://img.qammunity.org/2020/formulas/mathematics/college/oox1mxtzmqkef87gbrw1o1i0wzufdyhe0e.png)
So, Total cost of producing 5000 items is 48800
Now Average cost per item =
![\frac{\text{Total cost}}{\text{No. of items}}](https://img.qammunity.org/2020/formulas/mathematics/college/u4za6lmg4patn0qd421ldaa9mtucy3p659.png)
Average cost per item=
![(48800)/(5000)](https://img.qammunity.org/2020/formulas/mathematics/college/3hczjgkcxivh6w9aaz01qj0eg0ock5z2wi.png)
Average cost per item=9.76
So, the average cost per item when 5000 items are produced is 9.76