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

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1let 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

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

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

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

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

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.1hold 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

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