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cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
this is the product of 2 functions of L so use the product rule  use the Chain rule to differentiate each function
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
let u = 300L^0.8 du/dL = 240L^(0.2) let v = (240  5L)^0.5 dv?dL = 0.5(240  5L)^(0.5) * 5 = 2.5(240  5L)^(0.5) actually on the last function needed the chain rule Now plug in these values into the Product rule dQ/dL = u*dv/dL + v*du/dL
 one year ago

business1 Group TitleBest ResponseYou've already chosen the best response.0
like so: 300L^0.8*(2.5(2405L)^(0.5))+(2405L)^0.5*(240L^(0.2))?
 one year ago

business1 Group TitleBest ResponseYou've already chosen the best response.0
and if you were to equate this all to zero, then solve for L, you should end up with approximately 30. i cannot get there.
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
right  thats a horrible equation to solve using algebra. i'd be inclined to try to solve that graphically ( on a computer or graphical calculator)
 one year ago

business1 Group TitleBest ResponseYou've already chosen the best response.0
yeah so it seems. thanks very much.
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
hold on i've simplified it down to 750L*0.8 240(240  5L)^0.5  +  = 0 (240  5L)^0.5 L^0.2 multiply through by L^0.2 * (240  5L)^0.5 :  750L^ 0.8 * L^0.2 + 240 ( 240  5L) = 0 240^2  1200L  750L = 0 L = 240^2 / 1950 = 29.54
 one year ago

business1 Group TitleBest ResponseYou've already chosen the best response.0
this is just great! thank you very much.
 one year ago
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