he production function of a chocolate factory is given byQ = 100Le−0.25Lwhere L denotes the amount of labour and Q is the quantity of chocolate barsproduced.(a) Compute the first derivative dQdL .(b) Compute the second derivative d2QdL2 .(c) Find the value of L that maximizes the output Q
Question
he production function of a chocolate factory is given byQ = 100Le−0.25Lwhere L denotes the amount of labour and Q is the quantity of chocolate barsproduced.(a) Compute the first derivative dQdL .(b) Compute the second derivative d2QdL2 .(c) Find the value of L that maximizes the output Q
Solution
(a) To compute the first derivative dQ/dL, we need to apply the product rule and the chain rule of differentiation. The product rule is uv' + vu' and the chain rule is f(g(x))' = f'(g(x)) * g'(x).
The function can be rewritten as Q = 100L * e^(-0.25L).
So, u = 100L, v = e^(-0.25L), u' = 100, and v' can be found by applying the chain rule.
The derivative of e^x is e^x, and the derivative of -0.25L is -0.25. So, v' = e^(-0.25L) * -0.25.
Substituting these into the product rule gives:
dQ/dL = 100 * e^(-0.25L) + 100L * e^(-0.25L) * -0.25 = 100e^(-0.25L) - 25Le^(-0.25L)
(b) To compute the second derivative d²Q/dL², we need to differentiate dQ/dL with respect to L.
Again, we apply the product rule.
Let u = 100, v = e^(-0.25L), u' = 0, and v' = e^(-0.25L) * -0.25.
Then, let u = -25L, v = e^(-0.25L), u' = -25, and v' = e^(-0.25L) * -0.25.
Substituting these into the product rule gives:
d²Q/dL² = 0 - 25e^(-0.25L) - 25e^(-0.25L) + 6.25Le^(-0.25L) = -50e^(-0.25L) + 6.25Le^(-0.25L)
(c) To find the value of L that maximizes the output Q, we need to set the first derivative equal to zero and solve for L.
0 = 100e^(-0.25L) - 25Le^(-0.25L) = e^(-0.25L)(100 - 25L)
Setting each factor equal to zero gives:
e^(-0.25L) = 0 or 100 - 25L = 0
The exponential function is never zero, so the only solution is 100 - 25L = 0. Solving for L gives L = 4.
To confirm that this is a maximum, we can substitute L = 4 into the second derivative. If the second derivative is negative, then L = 4 is a maximum.
d²Q/dL² = -50e^(-0.254) + 6.254e^(-0.25*4) = -50e^(-1) + 25e^(-1) = -25e^(-1)
Since this is negative, L = 4 is indeed a maximum. So, the output Q is maximized when L = 4.
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