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business1

  • one year ago

Q=300L^0.8(240-5L)^0.5 dQ/dL=? thanks

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  1. cwrw238
    • one year ago
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    this is the product of 2 functions of L so use the product rule - use the Chain rule to differentiate each function

  2. cwrw238
    • one year ago
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    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

  3. business1
    • one year ago
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    like so: 300L^0.8*(-2.5(240-5L)^(-0.5))+(240-5L)^0.5*(240L^(-0.2))?

  4. cwrw238
    • one year ago
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    yes

  5. business1
    • one year ago
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    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.

  6. cwrw238
    • one year ago
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    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)

  7. business1
    • one year ago
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    yeah so it seems. thanks very much.

  8. cwrw238
    • one year ago
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    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

  9. business1
    • one year ago
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    this is just great! thank you very much.

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