A manufacturer has 750 meters of cotton and 1000 meters of polyester. Production of a sweatshirt requires 1 meter of cotton and 2 meters of polyester, while production of a shirt requires 1.5 meters of cotton and 1 meter of polyester. The sale prices of a sweatshirt and a shirt are 30 € and 24 €, respectively. What are the number of sweatshirts and the number of shirts that maximize total sales?
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S ≥ 0 C ≥ 0 S + 1 . 5 C ≤ 7 5 0 2 S + C ≤ 1 0 0 0
We have to maximize V ( S , C ) = 3 0 S + 2 4 C € and if we graphically represent on the cartesian plane the four before equations, where the horizontal axis is S and the vertical axis is C we obtain
For maximizing V ( S , C ) we only have to check this function at the vertices of the grey region(due to a fundamental theorem of linear programming )
V ( 0 , 0 ) = 0 € V ( 5 0 0 , 0 ) = 5 0 ⋅ 3 0 = 1 5 0 0 0 € V ( 0 , 5 0 0 ) = 5 0 0 ⋅ 2 4 = 1 2 0 0 0 € V ( 3 7 5 , 2 5 0 ) = 3 7 5 ⋅ 3 0 + 2 5 0 ⋅ 2 4 = 1 7 2 5 0 € .
Hence, the maximun number of sweatshirts S and maximum number of shirts C are 375 and 250 respectively, and therefore S + C = 6 2 5