## business1 2 years ago Q=300L^0.8(240-5L)^0.5 dQ/dL=? thanks

1. cwrw238

this is the product of 2 functions of L so use the product rule - use the Chain rule to differentiate each function

2. cwrw238

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

like so: 300L^0.8*(-2.5(240-5L)^(-0.5))+(240-5L)^0.5*(240L^(-0.2))?

4. cwrw238

yes

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

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)

yeah so it seems. thanks very much.

8. cwrw238

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