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Certainly! To solve this problem involving the total production function [tex]\( Y = 10x + 12x^2 - x^3 \)[/tex], we need to derive the average product (AP) and marginal product (MP) functions and prove that the marginal product function cuts the average product function at its maximum. Here is the step-by-step solution:
### Step 1: Total Production Function
Given:
[tex]\[ Y = 10x + 12x^2 - x^3 \][/tex]
### Step 2: Average Product (AP) Function
The average product function (AP) is calculated by dividing the total production [tex]\( Y \)[/tex] by the input [tex]\( x \)[/tex]:
[tex]\[ AP = \frac{Y}{x} = \frac{10x + 12x^2 - x^3}{x} \][/tex]
Simplify:
[tex]\[ AP = 10 + 12x - x^2 \][/tex]
So, the average product function is:
[tex]\[ AP = 10 + 12x - x^2 \][/tex]
### Step 3: Marginal Product (MP) Function
The marginal product function (MP) is the derivative of the total production function [tex]\( Y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ MP = \frac{dY}{dx} \][/tex]
Differentiate [tex]\( Y \)[/tex]:
[tex]\[ MP = \frac{d}{dx} (10x + 12x^2 - x^3) \][/tex]
Perform the differentiation:
[tex]\[ MP = 10 + 24x - 3x^2 \][/tex]
So, the marginal product function is:
[tex]\[ MP = 10 + 24x - 3x^2 \][/tex]
### Step 4: Finding Intersection Points
To find where the marginal product function cuts the average product function, we set [tex]\( AP \)[/tex] equal to [tex]\( MP \)[/tex]:
[tex]\[ 10 + 12x - x^2 = 10 + 24x - 3x^2 \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 12x - x^2 = 24x - 3x^2 \][/tex]
[tex]\[ -x^2 + 3x^2 = 24x - 12x \][/tex]
[tex]\[ 2x^2 = 12x \][/tex]
[tex]\[ 2x(x - 6) = 0 \][/tex]
So, the intersection points are:
[tex]\[ x = 0 \text{ or } x = 6 \][/tex]
### Step 5: Verifying Maximum of Average Product
To verify that the marginal product cuts the average product at its maximum, we need to check the second derivative of the average product function at the point of intersection:
[tex]\[ AP = 10 + 12x - x^2 \][/tex]
First derivative of [tex]\( AP \)[/tex]:
[tex]\[ AP' = 12 - 2x \][/tex]
Second derivative of [tex]\( AP \)[/tex]:
[tex]\[ AP'' = -2 \][/tex]
At [tex]\( x = 6 \)[/tex]:
[tex]\[ AP'' < 0 \][/tex]
Since the second derivative of [tex]\( AP \)[/tex] is negative at [tex]\( x = 6 \)[/tex], it confirms that [tex]\( AP \)[/tex] has a maximum at this point.
Thus, we have shown that the marginal product function cuts the average product function at its maximum value when [tex]\( x = 6 \)[/tex].
### Summary
- Total Production Function: [tex]\( Y = 10x + 12x^2 - x^3 \)[/tex]
- Average Product (AP) Function: [tex]\( AP = 10 + 12x - x^2 \)[/tex]
- Marginal Product (MP) Function: [tex]\( MP = 10 + 24x - 3x^2 \)[/tex]
- The marginal product cuts the average product at its maximum when [tex]\( x = 6 \)[/tex].
This completes the step-by-step solution.
### Step 1: Total Production Function
Given:
[tex]\[ Y = 10x + 12x^2 - x^3 \][/tex]
### Step 2: Average Product (AP) Function
The average product function (AP) is calculated by dividing the total production [tex]\( Y \)[/tex] by the input [tex]\( x \)[/tex]:
[tex]\[ AP = \frac{Y}{x} = \frac{10x + 12x^2 - x^3}{x} \][/tex]
Simplify:
[tex]\[ AP = 10 + 12x - x^2 \][/tex]
So, the average product function is:
[tex]\[ AP = 10 + 12x - x^2 \][/tex]
### Step 3: Marginal Product (MP) Function
The marginal product function (MP) is the derivative of the total production function [tex]\( Y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ MP = \frac{dY}{dx} \][/tex]
Differentiate [tex]\( Y \)[/tex]:
[tex]\[ MP = \frac{d}{dx} (10x + 12x^2 - x^3) \][/tex]
Perform the differentiation:
[tex]\[ MP = 10 + 24x - 3x^2 \][/tex]
So, the marginal product function is:
[tex]\[ MP = 10 + 24x - 3x^2 \][/tex]
### Step 4: Finding Intersection Points
To find where the marginal product function cuts the average product function, we set [tex]\( AP \)[/tex] equal to [tex]\( MP \)[/tex]:
[tex]\[ 10 + 12x - x^2 = 10 + 24x - 3x^2 \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 12x - x^2 = 24x - 3x^2 \][/tex]
[tex]\[ -x^2 + 3x^2 = 24x - 12x \][/tex]
[tex]\[ 2x^2 = 12x \][/tex]
[tex]\[ 2x(x - 6) = 0 \][/tex]
So, the intersection points are:
[tex]\[ x = 0 \text{ or } x = 6 \][/tex]
### Step 5: Verifying Maximum of Average Product
To verify that the marginal product cuts the average product at its maximum, we need to check the second derivative of the average product function at the point of intersection:
[tex]\[ AP = 10 + 12x - x^2 \][/tex]
First derivative of [tex]\( AP \)[/tex]:
[tex]\[ AP' = 12 - 2x \][/tex]
Second derivative of [tex]\( AP \)[/tex]:
[tex]\[ AP'' = -2 \][/tex]
At [tex]\( x = 6 \)[/tex]:
[tex]\[ AP'' < 0 \][/tex]
Since the second derivative of [tex]\( AP \)[/tex] is negative at [tex]\( x = 6 \)[/tex], it confirms that [tex]\( AP \)[/tex] has a maximum at this point.
Thus, we have shown that the marginal product function cuts the average product function at its maximum value when [tex]\( x = 6 \)[/tex].
### Summary
- Total Production Function: [tex]\( Y = 10x + 12x^2 - x^3 \)[/tex]
- Average Product (AP) Function: [tex]\( AP = 10 + 12x - x^2 \)[/tex]
- Marginal Product (MP) Function: [tex]\( MP = 10 + 24x - 3x^2 \)[/tex]
- The marginal product cuts the average product at its maximum when [tex]\( x = 6 \)[/tex].
This completes the step-by-step solution.
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