Answer :
Let's go through each part of this problem methodically, stepping through how to find the values that maximize output, marginal product, and average product, given the production function [tex]\( Q = 6L^2 - 0.4L^3 \)[/tex].
### Part A: Find the value of [tex]\( L \)[/tex] that maximizes output
To maximize the output [tex]\( Q \)[/tex], we need to find the critical points by setting the first derivative of [tex]\( Q \)[/tex] with respect to [tex]\( L \)[/tex] to zero. This involves finding [tex]\( \frac{dQ}{dL} \)[/tex].
1. Find the first derivative of [tex]\( Q \)[/tex]:
[tex]\[ \frac{dQ}{dL} = \frac{d}{dL}(6L^2 - 0.4L^3) = 12L - 1.2L^2 \][/tex]
2. Set the first derivative equal to zero to find critical points:
[tex]\[ 12L - 1.2L^2 = 0 \][/tex]
3. Solve the equation for [tex]\( L \)[/tex]:
[tex]\[ 1.2L^2 = 12L \][/tex]
[tex]\[ L(1.2L - 12) = 0 \][/tex]
[tex]\[ L = 0 \quad \text{or} \quad L = 10 \][/tex]
Thus, the values of [tex]\( L \)[/tex] that maximize output are [tex]\( L = 0 \)[/tex] and [tex]\( L = 10 \)[/tex].
### Part B: Find the value of [tex]\( L \)[/tex] that maximizes marginal product
To maximize the marginal product, we need to find the critical points of the marginal product function. The marginal product is the first derivative of the production function [tex]\( \frac{dQ}{dL} \)[/tex].
1. Marginal product function:
[tex]\[ MP = \frac{dQ}{dL} = 12L - 1.2L^2 \][/tex]
2. Find the derivative of the marginal product to determine where it is maximized:
[tex]\[ \frac{d(MP)}{dL} = 12 - 2.4L \][/tex]
3. Set the derivative equal to zero to find critical points:
[tex]\[ 12 - 2.4L = 0 \][/tex]
4. Solve the equation for [tex]\( L \)[/tex]:
[tex]\[ 2.4L = 12 \][/tex]
[tex]\[ L = 5 \][/tex]
Therefore, the value of [tex]\( L \)[/tex] that maximizes the marginal product is [tex]\( L = 5 \)[/tex].
### Part C: Find the value of [tex]\( L \)[/tex] that maximizes average product
To maximize the average product, we need to find the critical points of the average product function. The average product is defined as [tex]\( \frac{Q}{L} \)[/tex].
1. Average product function:
[tex]\[ AP = \frac{Q}{L} = \frac{6L^2 - 0.4L^3}{L} = 6L - 0.4L^2 \][/tex]
2. Find the first derivative of the average product:
[tex]\[ \frac{d(AP)}{dL} = 6 - 0.8L \][/tex]
3. Set the derivative equal to zero to find critical points:
[tex]\[ 6 - 0.8L = 0 \][/tex]
4. Solve the equation for [tex]\( L \)[/tex]:
[tex]\[ 0.8L = 6 \][/tex]
[tex]\[ L = 7.5 \][/tex]
Therefore, the value of [tex]\( L \)[/tex] that maximizes the average product is [tex]\( L = 7.5 \)[/tex].
### Summary
- The values of [tex]\( L \)[/tex] that maximize output ([tex]\( Q \)[/tex]): [tex]\( L = 0 \)[/tex] and [tex]\( L = 10 \)[/tex]
- The value of [tex]\( L \)[/tex] that maximizes the marginal product: [tex]\( L = 5 \)[/tex]
- The value of [tex]\( L \)[/tex] that maximizes the average product: [tex]\( L = 7.5 \)[/tex]
### Part A: Find the value of [tex]\( L \)[/tex] that maximizes output
To maximize the output [tex]\( Q \)[/tex], we need to find the critical points by setting the first derivative of [tex]\( Q \)[/tex] with respect to [tex]\( L \)[/tex] to zero. This involves finding [tex]\( \frac{dQ}{dL} \)[/tex].
1. Find the first derivative of [tex]\( Q \)[/tex]:
[tex]\[ \frac{dQ}{dL} = \frac{d}{dL}(6L^2 - 0.4L^3) = 12L - 1.2L^2 \][/tex]
2. Set the first derivative equal to zero to find critical points:
[tex]\[ 12L - 1.2L^2 = 0 \][/tex]
3. Solve the equation for [tex]\( L \)[/tex]:
[tex]\[ 1.2L^2 = 12L \][/tex]
[tex]\[ L(1.2L - 12) = 0 \][/tex]
[tex]\[ L = 0 \quad \text{or} \quad L = 10 \][/tex]
Thus, the values of [tex]\( L \)[/tex] that maximize output are [tex]\( L = 0 \)[/tex] and [tex]\( L = 10 \)[/tex].
### Part B: Find the value of [tex]\( L \)[/tex] that maximizes marginal product
To maximize the marginal product, we need to find the critical points of the marginal product function. The marginal product is the first derivative of the production function [tex]\( \frac{dQ}{dL} \)[/tex].
1. Marginal product function:
[tex]\[ MP = \frac{dQ}{dL} = 12L - 1.2L^2 \][/tex]
2. Find the derivative of the marginal product to determine where it is maximized:
[tex]\[ \frac{d(MP)}{dL} = 12 - 2.4L \][/tex]
3. Set the derivative equal to zero to find critical points:
[tex]\[ 12 - 2.4L = 0 \][/tex]
4. Solve the equation for [tex]\( L \)[/tex]:
[tex]\[ 2.4L = 12 \][/tex]
[tex]\[ L = 5 \][/tex]
Therefore, the value of [tex]\( L \)[/tex] that maximizes the marginal product is [tex]\( L = 5 \)[/tex].
### Part C: Find the value of [tex]\( L \)[/tex] that maximizes average product
To maximize the average product, we need to find the critical points of the average product function. The average product is defined as [tex]\( \frac{Q}{L} \)[/tex].
1. Average product function:
[tex]\[ AP = \frac{Q}{L} = \frac{6L^2 - 0.4L^3}{L} = 6L - 0.4L^2 \][/tex]
2. Find the first derivative of the average product:
[tex]\[ \frac{d(AP)}{dL} = 6 - 0.8L \][/tex]
3. Set the derivative equal to zero to find critical points:
[tex]\[ 6 - 0.8L = 0 \][/tex]
4. Solve the equation for [tex]\( L \)[/tex]:
[tex]\[ 0.8L = 6 \][/tex]
[tex]\[ L = 7.5 \][/tex]
Therefore, the value of [tex]\( L \)[/tex] that maximizes the average product is [tex]\( L = 7.5 \)[/tex].
### Summary
- The values of [tex]\( L \)[/tex] that maximize output ([tex]\( Q \)[/tex]): [tex]\( L = 0 \)[/tex] and [tex]\( L = 10 \)[/tex]
- The value of [tex]\( L \)[/tex] that maximizes the marginal product: [tex]\( L = 5 \)[/tex]
- The value of [tex]\( L \)[/tex] that maximizes the average product: [tex]\( L = 7.5 \)[/tex]
