Final Answer:
(a) The isoquant for the production function f(x1, x2) = min{x1, 2x2} appears as a straight line with a slope of 1/2, showing the trade-off between copper (x1) and zinc (x2).
(b) This production function exhibits decreasing returns to scale.
(c) To produce 10 effronteries, the firm would require 10 units of copper and 5 units of zinc.
(d) When facing factor prices (1, 1), the cheapest way to produce 10 effronteries is by using 10 units of copper and 5 units of zinc, costing $15.
(e) The cheapest cost to produce 10 effronteries when facing factor prices (w1, w2) is w1*10 + w2*5.
(f) When facing factor prices (w1, w2), the minimal cost of producing Y effronteries is w1*Y + (w2/2)*Y.
Step-by-step explanation:
(a) The isoquant depicts the combinations of inputs (copper and zinc) that produce the same level of output. For f(x1, x2) = min{x1, 2x2}, the isoquant is a straight line with a slope of 1/2, representing the fixed proportions of the alloy.
(b) Decreasing returns to scale occur when an increase in inputs leads to a proportionately smaller increase in output. In this case, doubling both inputs more than doubles the output, indicating decreasing returns to scale.
(c) To produce 10 effronteries, the production function's minimum requirement is 10 units of copper (x1) and 5 units of zinc (x2), based on the function f(x1, x2) = min{x1, 2x2}.
(d) When facing factor prices (1, 1), the firm minimizes costs by using the input combination that satisfies the production function: 10 units of copper and 5 units of zinc, resulting in a total cost of $15 (1*10 + 1*5).
(e) With factor prices (w1, w2), the cheapest cost to produce 10 effronteries is determined by the total cost equation: w1*10 + w2*5.
(f) Extending this logic for Y effronteries and factor prices (w1, w2), the minimal cost becomes w1*Y + (w2/2)*Y, as the production function's fixed proportionality necessitates half the units of zinc compared to copper.