# 10.6.3. Linear Programing and Optimization

A farmer found that he can improve the quality of his produce if he uses at least 18 units of nitrogen-based Compounds and 6 units of phosphate Compounds for each hectare. There are two types of fertilizers; and . The contents and cost of each are shown in the following table:

The FertilizerNumber of Units of Nitrogen-Based Compounds per kgNumber of Units of Phosphate Compounds per kgCost for each kg (LE)
A32170
B61120

Given that the following figure represents this situation, find the least cost of combination of fertilizers and so that the farmer can provide a sufficient number of units of both compounds to improve the quality of his produce.

• ALE
• BLE
• CLE
• DLE

### Example

A farmer found that he can improve the quality of his produce if he uses at least 18 units of nitrogen-based Compounds and 6 units of phosphate Compounds for each hectare. There are two types of fertilizers; and . The contents and cost of each are shown in the following table:

The FertilizerNumber of Units of Nitrogen-Based Compounds per kgNumber of Units of Phosphate Compounds per kgCost for each kg (LE)
A32170
B61120

Given that the following figure represents this situation, find the least cost of combination of fertilizers and so that the farmer can provide a sufficient number of units of both compounds to improve the quality of his produce.

### Solution

Since the price of a kilogram of fertilizer A is LE, the total price paid by the farmer for fertilizer A is LE, and since the price of a kilogram of fertilizer B is LE, the total price paid by the farmer for fertilizer B is LE. Therefore, if the total price of all the fertilizer is , the objective function is . The vertices of the feasible region are , , and . First, find the value of the objective function at the vertex .

Next, find the value of the objective function at the vertex .

Finally, find the value of the objective function at the vertex .

Thus, finding the value of the objective function at each vertex shows that the minimum value of the function is LE, so this is the least cost of the fertilizers for the farmer to improve the quality of his produce.

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