Answer :
Let's work through each part of the problem step-by-step.
### Part a) Find the value of [tex]\( L \)[/tex] that maximizes output
Given the production function:
[tex]\[ Q = 6L^2 - 0.4L^3, \][/tex]
we want to find the value of [tex]\( L \)[/tex] that maximizes the output [tex]\( Q \)[/tex].
First, we need to find the first derivative of [tex]\( Q \)[/tex] with respect to [tex]\( L \)[/tex]:
[tex]\[ \frac{dQ}{dL} = 12L - 1.2L^2. \][/tex]
To find the critical points, we set the first derivative equal to zero and solve for [tex]\( L \)[/tex]:
[tex]\[ 12L - 1.2L^2 = 0 \][/tex]
[tex]\[ L(12 - 1.2L) = 0 \][/tex]
[tex]\[ L = 0 \quad \text{or} \quad L = 10. \][/tex]
Next, we check the second derivative of [tex]\( Q \)[/tex] to determine if these critical points are maxima or minima:
[tex]\[ \frac{d^2Q}{dL^2} = 12 - 2.4L. \][/tex]
Evaluating the second derivative at [tex]\( L = 10 \)[/tex]:
[tex]\[ \frac{d^2Q}{dL^2} \bigg|_{L=10} = 12 - 2.4 \cdot 10 = 12 - 24 = -12. \][/tex]
Since the second derivative is negative at [tex]\( L = 10 \)[/tex], it indicates a maximum. Thus, the value of [tex]\( L \)[/tex] that maximizes output is:
[tex]\[ L = 10. \][/tex]
### Part b) Find the value of [tex]\( L \)[/tex] that maximizes marginal product
The marginal product (MP) is the first derivative of the production function:
[tex]\[ MP = \frac{dQ}{dL} = 12L - 1.2L^2. \][/tex]
To maximize the marginal product, we need to find the critical points of its derivative:
[tex]\[ \frac{d(MP)}{dL} = 12 - 2.4L. \][/tex]
Set it equal to zero:
[tex]\[ 12 - 2.4L = 0 \][/tex]
[tex]\[ L = 5. \][/tex]
Next, we check the second derivative to confirm it’s a maximum:
[tex]\[ \frac{d^2(MP)}{dL^2} = -2.4. \][/tex]
Since the second derivative is negative, it indicates a maximum. Thus, the value of [tex]\( L \)[/tex] that maximizes marginal product is:
[tex]\[ L = 5. \][/tex]
### Part c) Find the value of [tex]\( L \)[/tex] that maximizes average product
The average product (AP) is given by the ratio of output to the input:
[tex]\[ AP = \frac{Q}{L} = \frac{6L^2 - 0.4L^3}{L} = 6L - 0.4L^2. \][/tex]
To maximize the average product, we take the derivative of AP with respect to [tex]\( L \)[/tex] and find its critical points:
[tex]\[ \frac{d(AP)}{dL} = 6 - 0.8L. \][/tex]
Set it equal to zero:
[tex]\[ 6 - 0.8L = 0 \][/tex]
[tex]\[ L = 7.5. \][/tex]
Next, we confirm it’s a maximum by checking the second derivative:
[tex]\[ \frac{d^2(AP)}{dL^2} = -0.8. \][/tex]
Since the second derivative is negative, it indicates a maximum. Thus, the value of [tex]\( L \)[/tex] that maximizes average product is:
[tex]\[ L = 7.5. \][/tex]
### Summary of Results
- The value of [tex]\( L \)[/tex] that maximizes output is [tex]\( L = 10 \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes marginal product is [tex]\( L = 5 \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes average product is [tex]\( L = 7.5 \)[/tex].
### Part a) Find the value of [tex]\( L \)[/tex] that maximizes output
Given the production function:
[tex]\[ Q = 6L^2 - 0.4L^3, \][/tex]
we want to find the value of [tex]\( L \)[/tex] that maximizes the output [tex]\( Q \)[/tex].
First, we need to find the first derivative of [tex]\( Q \)[/tex] with respect to [tex]\( L \)[/tex]:
[tex]\[ \frac{dQ}{dL} = 12L - 1.2L^2. \][/tex]
To find the critical points, we set the first derivative equal to zero and solve for [tex]\( L \)[/tex]:
[tex]\[ 12L - 1.2L^2 = 0 \][/tex]
[tex]\[ L(12 - 1.2L) = 0 \][/tex]
[tex]\[ L = 0 \quad \text{or} \quad L = 10. \][/tex]
Next, we check the second derivative of [tex]\( Q \)[/tex] to determine if these critical points are maxima or minima:
[tex]\[ \frac{d^2Q}{dL^2} = 12 - 2.4L. \][/tex]
Evaluating the second derivative at [tex]\( L = 10 \)[/tex]:
[tex]\[ \frac{d^2Q}{dL^2} \bigg|_{L=10} = 12 - 2.4 \cdot 10 = 12 - 24 = -12. \][/tex]
Since the second derivative is negative at [tex]\( L = 10 \)[/tex], it indicates a maximum. Thus, the value of [tex]\( L \)[/tex] that maximizes output is:
[tex]\[ L = 10. \][/tex]
### Part b) Find the value of [tex]\( L \)[/tex] that maximizes marginal product
The marginal product (MP) is the first derivative of the production function:
[tex]\[ MP = \frac{dQ}{dL} = 12L - 1.2L^2. \][/tex]
To maximize the marginal product, we need to find the critical points of its derivative:
[tex]\[ \frac{d(MP)}{dL} = 12 - 2.4L. \][/tex]
Set it equal to zero:
[tex]\[ 12 - 2.4L = 0 \][/tex]
[tex]\[ L = 5. \][/tex]
Next, we check the second derivative to confirm it’s a maximum:
[tex]\[ \frac{d^2(MP)}{dL^2} = -2.4. \][/tex]
Since the second derivative is negative, it indicates a maximum. Thus, the value of [tex]\( L \)[/tex] that maximizes marginal product is:
[tex]\[ L = 5. \][/tex]
### Part c) Find the value of [tex]\( L \)[/tex] that maximizes average product
The average product (AP) is given by the ratio of output to the input:
[tex]\[ AP = \frac{Q}{L} = \frac{6L^2 - 0.4L^3}{L} = 6L - 0.4L^2. \][/tex]
To maximize the average product, we take the derivative of AP with respect to [tex]\( L \)[/tex] and find its critical points:
[tex]\[ \frac{d(AP)}{dL} = 6 - 0.8L. \][/tex]
Set it equal to zero:
[tex]\[ 6 - 0.8L = 0 \][/tex]
[tex]\[ L = 7.5. \][/tex]
Next, we confirm it’s a maximum by checking the second derivative:
[tex]\[ \frac{d^2(AP)}{dL^2} = -0.8. \][/tex]
Since the second derivative is negative, it indicates a maximum. Thus, the value of [tex]\( L \)[/tex] that maximizes average product is:
[tex]\[ L = 7.5. \][/tex]
### Summary of Results
- The value of [tex]\( L \)[/tex] that maximizes output is [tex]\( L = 10 \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes marginal product is [tex]\( L = 5 \)[/tex].
- The value of [tex]\( L \)[/tex] that maximizes average product is [tex]\( L = 7.5 \)[/tex].
