Linear programming

Algebra Level 2

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 ( S ) (S) and the number of shirts ( C ) (C) that maximize total sales?

Submit S + C . S + C.


The answer is 625.

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1 solution

S 0 C 0 S + 1.5 C 750 2 S + C 1000 S \ge 0 \\ C \ge 0 \\ S + 1.5C \leq 750 \\ 2S + C \leq 1000

We have to maximize V ( S , C ) = 30 S + 24 C V(S,C) = 30S + 24C \space € and if we graphically represent on the cartesian plane the four before equations, where the horizontal axis is S S and the vertical axis is C C we obtain

For maximizing V ( S , C ) 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 ( 500 , 0 ) = 50 30 = 15000 V ( 0 , 500 ) = 500 24 = 12000 V ( 375 , 250 ) = 375 30 + 250 24 = 17250 V(0,0) = 0 \space € \\ V(500,0) = 50 \cdot 30 = 15000 \space € \\ V(0,500) = 500 \cdot 24 =12000 \space € \\ V(375,250) = 375 \cdot 30 + 250 \cdot 24 = 17250 \space € .

Hence, the maximun number of sweatshirts S S and maximum number of shirts C C are 375 and 250 respectively, and therefore S + C = 625 S + C = 625

And nice solution sir.

Samara Simha Reddy - 5 years, 2 months ago

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Please, I know I'm not young, but don't call me sir, please.... Anyway, thank you, I think soon there will be more people with right answer,There are 11 views, 3 attempts and 1 solver(you), so far... We'll see later.

Guillermo Templado - 5 years, 2 months ago

I think this is easy sum. But no solvers. I m the first to get this problem right.

Samara Simha Reddy - 5 years, 2 months ago

Thank you! Could you share how you draw these graphs?

Gustavo Rangel - 7 months, 2 weeks ago

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