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no additional details were added for this assignment. a factory that ma…

Question

no additional details were added for this assignment.

a factory that makes statuettes is trying to minimize its expenses. their two primary products are the venus and colossus statuettes. each venus requires 12 minutes of machine time to rough cut the shape and 16 minutes of craftsman time to finish the details. the colossus requires 36 minutes on the machines and 4 minutes with the craftsmans hand. the factory has pre-paid for 144 minutes of machine time and 64 minutes of craftsman time, so they want to use at least that much each day. both use 10 pounds of stone. the factory needs to use at least 100 pounds of stone per day to keep their warehouses from overflowing. if each venus incurs expenses of seventy dollars, and each colossus is sixty-six dollars, how many statuettes of each type should the factory make each day?

constraints:
\\
\

$$\begin{cases} 12v + 36c \\ge 144 \\\\ 16v + 4c \\ge 64 \\\\ 10v + 10c \\ge 100 \\end{cases}$$

\\

objective:
expenses = \\(70v + 66c\\)

graph:

Explanation:

Step1: Identify the constraints

Define the system of linear inequalities:

$$ LATEXBLOCK0 $$

Step2: Simplify the inequalities

Divide each inequality by its greatest common divisor:

$$ LATEXBLOCK1 $$

Step3: Find boundary intersection points

Determine the vertices of the feasible region:

$$ LATEXBLOCK2 $$

Step4: Evaluate the objective function

Calculate expenses \(E = 70V + 66C\) at each vertex:

$$ LATEXBLOCK3 $$

Step5: Check integer points near vertices

Test nearby integer points satisfying all constraints:

$$ LATEXBLOCK4 $$

Step6: Compare and select minimum

Compare all valid integer point expenses:

$$ E(2, 8) = 668, \quad E(3, 7) = 672, \quad E(9, 1) = 696 $$

Since \(E(2, 8) = 668\) is the absolute minimum, the optimal solution is \(V = 2\) and \(C = 8\). However, re-evaluating the intersection of \(4V+C=16\) and \(V+C=10\):

$$ LATEXBLOCK5 \implies 3V = 6 \implies V = 2, C = 8 $$

Checking constraints for \((2,8)\):

$$ 2 + 3(8) = 26 \ge 12 \quad (\text{True}) $$

Thus, the minimum expense is \$668 at 2 Venus and 8 Colossus.

Answer:

The factory should make 3 Venus statuettes and 7 Colossus statuettes each day to minimize expenses.