Answer:
The minimum cost will be "$214085".
Step-by-step explanation:
![D = 1700 units \\\\S = \$ 50 \\\\H= 20%\\](https://img.qammunity.org/2021/formulas/business/college/g1yuwx7rhtxdzonugihhrpfrrcorbmc4bn.png)
i) When quantity = 1-1500, price = $ 12.50 , and holding price is $12.50 * 20 %= $2.50.
ii) When quantity = 1501 -10,000, price = $ 12.45 , and holding price is $12.45 * 20 %= $2.49.
iii) When quantity = 10,0001- and more, price = $ 12.40 , and holding price is $12.40 * 20 %= $2.48.
![EOQ= \sqrt{(2DS)/(H)} \\\\EOQ1= \sqrt{(2* 17000* 50)/(2.50)} \\\\EOQ1=824.62 \ \ \ or \ \ \ 825\\](https://img.qammunity.org/2021/formulas/business/college/6un4o9hraqfpv8tk1wie1k9209qunxq0wt.png)
![EOQ2= \sqrt{(2* 17000* 50)/(2.49)} \\\\EOQ1=826.2T \ \ \ or \ \ \ 826\\](https://img.qammunity.org/2021/formulas/business/college/mibemq8idtmsn3yajmso3s7e93t02l5jzs.png)
![EOQ3= \sqrt{(2* 17000* 50)/(2.48)} \\\\EOQ3=827.93 \ \ \ or \ \ \ 828\\](https://img.qammunity.org/2021/formulas/business/college/ays19hvalz67dqrqvjszs6e50xwpfztj5l.png)
know we should calculate the total cost of EOQ1 and break ever points (1501 to 10,000)units
![total \ cost = odering \ cost + holding \ cost + \ Annual \ product \ cost\\\\total_c = (D)/(Q) * S + (Q)/(2) * H + (p * D) \\\\T_c = (17000)/(825) * 50+ (825)/(2) * 2.50 + (12.50 * 17000)\\\\T_c = 1030 .30 +1031.25+212500\\\\T_c =$ 214561.55\\\\](https://img.qammunity.org/2021/formulas/business/college/nm42jxdmvbyl37qhnwn1dda08y4acgsvki.png)
![T_c = (17000)/(1501) * 50+ (1501)/(2) * 2.49 + (12.45 * 17000)\\\\T_c = 566.28 +1868.74+211650\\\\T_c =$ 214085.02 \ \ \ or \ \ \ $ 214085\\\\](https://img.qammunity.org/2021/formulas/business/college/neh7gzkazcegawpkg8qs21542trcldqcxc.png)
![T_c = (17000)/(10001) * 50+ (10001)/(2) * 2.48 + (12.40 * 17000)\\\\T_c = 84.99+ 12401.24+210800\\\\T_c =$ 223286.23 \\](https://img.qammunity.org/2021/formulas/business/college/b27tb8hpo4t2qvzplwld34r584bl3p1g74.png)
The total cost is less then 15001. So, optimal order quantity is 1501, that's why cost is = $214085.