Question

Suppose you are running a factory, producing anvils that requires steel as a raw material. Your costs are predominantly human labor (h), which is $20 per hour for your workers, and the steel itself (s), which runs for $160 per ton. Suppose your revenue is loosely modeled by the following equation: 100h^(2/3) s^(1/3). If your budget is $10,000, how would you maximize revenue? Solve this problem using the Lagrangian method.

Answer 1

Answer: - In the question we have been Given, The Revenue Function Price of Human Labor (h)= $20 Price of Steel (s) = $160 look at Total cost with the Producce = 10,000 we need to maximise the Cost Constraint the revenue function given TC =Pnxh +Ps x S 10000 20h + 160S So, we will use the lagrangian Methal as this is constrained Maximisation Problem ise Economic theory says maximise revenue till our Cost Bb Production allows: as we can looh So, we Express the Maximisation Problem Rch,s) = Subject to 10000 20h + 16os (from ) into Lagrangian function (d) L= + (10,000 Revenue function lagrangian The Cost of Constraint. multiplier (A Constrant M 1 looh si •20h-160s)

For the maximisection of revenue it is that necessary Practical derivative d 2 with respect to his and I do above lagrangian function be equal to Zero dk d dh ad го co th ds 2 x 100th s² - 20d=0 > d=10 (25) de - 33 x 100 hs s' _ 1600 =0 > d= (s) - (111) Comparing (11) and Cul) d = io (8/4)' = 5 h/s)? he (Divided both side 5294%) 48 = 485 (W 10,000 - 20h - 1605 dd d - 20h +160s. = 10,000 → 20 (489) + 1605 = 10,000 from A 11 los 10,000 8.928 ton h = 1428.588 hours)