Answer:
Minimum cost = 1200$
Step-by-step explanation:
We are given the following information in the question:
An animal food company must produce 200 kg

No more than 80 kg Of first ingredient can be used and at
least 60 kg Of second ingredient must be used.

Cost of ingredient
= Rs 3 per kg
Cost of ingredient
= Rs 8 per kg
Total cost = 
We have to minimize this cost.
Then, we can write the following inequalities:

The corner points as evaluated from graph are: (0,200) and (80,120).
C(0,200) = 1600$
C(80,120) = 1200$
Hence, by corner point theorem, the minimum cost would be 1200$ when 80 kg of first ingredient is used and 120 kg of second ingredient.
The attached image shows the graph.