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Math  /  Geometry

QuestionMinimize: z=700x+600y\quad z=700 x+600 y

Studdy Solution

STEP 1

1. The problem is a linear programming problem.
2. The objective function to minimize is z=700x+600y z = 700x + 600y .
3. The feasible region is defined by the vertices: (5,45) (5, 45) , (40,45) (40, 45) , (30,40) (30, 40) , (35,25) (35, 25) , and (40,5) (40, 5) .

STEP 2

1. Identify the vertices of the feasible region.
2. Evaluate the objective function at each vertex.
3. Determine the minimum value of the objective function.

STEP 3

Identify the vertices of the feasible region:
The vertices are given as: - (5,45) (5, 45) - (40,45) (40, 45) - (30,40) (30, 40) - (35,25) (35, 25) - (40,5) (40, 5)

STEP 4

Evaluate the objective function z=700x+600y z = 700x + 600y at each vertex:
1. At (5,45) (5, 45) : $ z = 700(5) + 600(45) = 3500 + 27000 = 30500 \]
2. At (40,45) (40, 45) : $ z = 700(40) + 600(45) = 28000 + 27000 = 55000 \]
3. At (30,40) (30, 40) : $ z = 700(30) + 600(40) = 21000 + 24000 = 45000 \]
4. At (35,25) (35, 25) : $ z = 700(35) + 600(25) = 24500 + 15000 = 39500 \]
5. At (40,5) (40, 5) : $ z = 700(40) + 600(5) = 28000 + 3000 = 31000 \]

STEP 5

Determine the minimum value of the objective function:
The calculated values are: - z=30500 z = 30500 at (5,45) (5, 45) - z=55000 z = 55000 at (40,45) (40, 45) - z=45000 z = 45000 at (30,40) (30, 40) - z=39500 z = 39500 at (35,25) (35, 25) - z=31000 z = 31000 at (40,5) (40, 5)
The minimum value is 30500 \boxed{30500} at the vertex (5,45) (5, 45) .

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