Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

yes

Answer 5

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.

Answer 6

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)

Answer 7

yeah so it seems. thanks very much.

Answer 8

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

Answer 9

this is just great! thank you very much.