Suppose a company produces a commodity X using an input L. The production function is X =
100L – L2.
(a.) Find what level of L maximizes production output, X.
(b.) What levels of L generate an output of X = 0? (there are two different L values)
Suppose a company produces a commodity X using an input L. The production function is X =
100L – L2.
(a.) Find what level of L maximizes production output, X.
(b.) What levels of L generate an output of X = 0? (there are two different L values)
Do you mean that your production function is $\displaystyle X=100L-L^2$ or $\displaystyle X=100L - 2L$?
I suspect it is the former. If you know derivatives, you can take the derivative and set it equal to zero. If not, you need to complete the square and the maximum will be at the vertex since this is a downward facing parabola.
$\displaystyle X=100L-L^2=-(L^2-100L)$
$\displaystyle X=-((L-50)^2-50^2)$
$\displaystyle X=-(L-50)^2+50^2$
Thus the vertex is at $\displaystyle (50,50^2)$
So for part a, X is maximized when L=50
For part b solve $\displaystyle X=0=-(L-50)^2+50^2$
$\displaystyle 50^2=(L-50)^2$
$\displaystyle \pm50=L-50$
$\displaystyle L=0,100$
Alternatively you could have solved
$\displaystyle X=0=100L-L^2=L(100-L)$ which yields the same two answers for L.