Video Transcript
A seafood restaurant sells two types of cooked fish: cod and eel. The restaurant sells no less than 40 fish every day. But it does not use more than 30 cod and no more than 45 eels. The price of one cod is six Egyptian pounds, and that of an eel is eight Egyptian pounds. Let 𝑥 represent the amount of cod purchased each day and 𝑦 represent the amount of eel. Given that the manager wants to minimize the total price 𝑝 of fish, state the objective function and the inequalities that will help the restaurant manager decide how many of each fish to buy.
This is an example of a linear programming problem. We are asked to state the objective function which can be used to optimize a quantity. In this case, the restaurant manager wants to minimize the total price of fish. We are also asked to state the inequalities or constraints which will help him do this.
We are told to let 𝑥 represent the amount of cod purchased and 𝑦 represent the amount of eel. Since neither of these can be negative, our first two inequalities are 𝑥 is greater than or equal to zero and 𝑦 is greater than or equal to zero. We are told that the restaurant sells no less than 40 fish. This means that the total number of fish purchased must be greater than or equal to 40. We have the inequality 𝑥 plus 𝑦 is greater than or equal to 40. The restaurant does not use more than 30 cod. This means that 𝑥 must be less than or equal to 30. They also do not use more than 45 eels. So 𝑦 must be less than or equal to 45. These five inequalities will help the restaurant manager decide how many of each fish to buy.
Our final step is to find an expression for the total price 𝑝. We know that the price of one cod is six Egyptian pounds and the price of one eel is eight Egyptian pounds. The price 𝑝 is therefore equal to six 𝑥 plus eight 𝑦. This is the objective function that the manager needs to find the minimum value of while satisfying the five inequalities.
Whilst it is not required in this question, one way of doing this is to sketch a graph of the inequalities and find the feasible region. The optimal solution, in this case the minimum price, will be found at one of the vertices of this feasible region.
