Question 12.14: In the theory of individual labour supply it is assumed that......

Given a maximum working day of 12 hours, then hours of work H = 12 − L.
Therefore, given an hourly wage of £4, income earned will be

I = wH = w(12 − L) = 4(12 − L) = 48 − 4L        (1)

Substituting (1) into the utility function

U = L^{0.5}I^{0.75} = L^{0.5}(48 − 4L)^{0.75}          (2)

To differentiate U using the product rule let

u = L^{0.5}   and    v = (48 − 4L) ^{0.75}

giving

\frac{du}{dL} = 0.5 L^{-0.5}         \frac{dv}{dL} = 0.75(48 − 4L) ^{-0.25}(−4)
                                       = −3(48 − 4L) ^{-0.25}

Therefore

\frac{dU}{dL} = L^{0.5}[−3(48 − 4L)^{-0.25}] + (48 − 4L) ^{0.75}(0.5 L^{-0.5})
= \frac{- 3L  +  (48  −  4L)0.5}{(48  −  4L)^{0.25}L^{0.5}}
= \frac{24  –  5L}{(48  −  4L)^{0.25}L^{0.5}} = 0              (3)

for a stationary point. Therefore

24 − 5L = 0
24 = 5L
4.8 = L

and so

H = 12 − 4.8 = 7.2 hours

To check the second-order condition we need to differentiate (3) again. Let

u = 24 − 5L     and     v = (48 − 4L)^{0.25} L^{0.5}

giving

\frac{du}{dL} = −5

and

\frac{dv}{dL}  =  (48 − 4L) ^{0.25} 0.5 L^{-0.5} +  L^{0.5} 0.25(48 − 4L) ^{-0.75}(−4)
= \frac{(48  −  4L)0.5  −  L}{L^{0.5}(48  −  4L)^{0.75}}
= \frac{24  –  3L}{L^{0.5}(48  −  4L)^{0.75}}

Therefore, using the quotient rule,

\frac{d^2U}{dL^2}  =  \frac{(48  −  4L)^{0.25}L^{0.5}(−5)  − (24  −  5L)[(24  −  3L)/L^{0.5}(48  −  4L)^{0.75}]}{(48  −  4L)^{0.5}L}

When L = 4.8 then 24 − 5L = 0 and so the second part of the numerator disappears. Then, dividing through top and bottom by (48 − 4L) ^{0.25}L^{0.5} we get

\frac{d^2U}{dL^2} = \frac{-5}{(48 − 4L)^{0.25}L^{0.5}}  = −0.985 < 0

and so the second-order condition for maximization of utility is satisfied when 7.2 hours are worked and 4.8 hours are taken as leisure.