Consider the following short run production function, `Q=100L-L^2,` where `Q` is the output level and `L` is Labour input. If the price of output...

Hello!

A profit is an income minus an expense. In this problem the expense is `1200L` and the income is `50(100L-l^2).` Therefore the profit as a function of `L` is

`P=P(L)=50(100L-L^2)-1200L=50(100L-L^2-24L)=`

`=50(76L-L^2).`

The graph of this function is a parabola branches down, it reaches its maximum at the vertex. The `L`-coordinate of the vertex is `L=76/2=38,` and the `P` coordinate is  `50*38*38=72,200.`

The output level is  `38*38=1,444.`

(we can also obtain the same result using differentiation, `P'(L)=0`)

So the answers are: profit is maximized if the firm uses 38 hours of labour, the output level is 1,444, and the maximum profit is 72,200 K.sh.