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business1
 2 years ago
Q=300L^0.8(2405L)^0.5
dQ/dL=?
thanks
business1
 2 years ago
Q=300L^0.8(2405L)^0.5 dQ/dL=? thanks

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cwrw238
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0yeah so it seems. thanks very much.

cwrw238
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0this is just great! thank you very much.
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