Profit Maximization Analysis with Milk Price Variation
Using a milk price of $10/cwt, grain price of $3/cwt, and hay price of $1.50/cwt, the profit-maximizing inputs for milk production are determined through mathematical calculations. The optimal weekly inputs are found to be 182.5 lbs of grain and 118.2 lbs of hay, resulting in a weekly profit of $23.35.
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Think Break #8 Repeat the previous problem, but now use a milk price of Milk $10/cwt M = 25.9 + 2.56G + 1.05H 0.00505G2 0.00109H2 0.00352GH Prices: Grain $3/cwt, Hay $1.50/cwt What are the profit maximizing inputs?
Think Break #8: Answer = 0.1( 25.9 + 2.56G + 1.05H 0.00505G2 0.00109H2 0.00352GH) 0.03G 0.015H FOC s G= 0.1(2.56 0.0101G 0.00352H) 0.03 = 0 H= 0.1(1.05 0.00218H 0.00352G) 0.015 = 0 2.56 0.0101G 0.00352H = 0.03/0.10 = 0.3 1.05 0.00218H 0.00352G = 0.015/0.10 = 0.15
Think Break #8: Answer Solve FOC1 for H 2.56 0.0101G 0.00352H = 0.3 0.00352H = 2.26 0.0101G H = (2.26 0.0101G)/0.00352 H = 642 2.87G Substitute into FOC2 and solve for G 1.05 0.00218H 0.00352G = 0.15 0.9 = 0.00218(642 2.87G) + 0.00352G 0.9 = 1.4 0.00626G + 0.00352G 0.5 = 0.00274G G = 0.5/0.00274 = 182.5 lbs/week
Think Break #8: Answer Substitute into equation for H H = 642 2.87G H = 642 2.87(182.5) H = 642 523.8 = 118.2 lbs/week M = 25.9 + 2.56G + 1.05H 0.00505G2 0.00109H2 0.00352GH M = 25.9 + 2.56(182.5) + 1.05(118.2) 0.00505(182.5)2 0.00109(118.2)2 0.00352(182.5)(118.2) = 306 lbs/week = 0.1(306) 0.03(182.5) 0.015(118.2) = $23.35/week
