Answer :
Certainly! To determine [tex]\( P_x(1400, 1800) \)[/tex] and [tex]\( P_y(1400, 1800) \)[/tex], we first need to define what [tex]\( P_x \)[/tex] and [tex]\( P_y \)[/tex] are. [tex]\( P_x \)[/tex] and [tex]\( P_y \)[/tex] denote the partial derivatives of profit with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex] respectively, where profit [tex]\( P \)[/tex] is defined as the difference between revenue [tex]\( R \)[/tex] and cost [tex]\( C \)[/tex].
Given the functions for revenue [tex]\( R(x, y) \)[/tex] and cost [tex]\( C(x, y) \)[/tex],
[tex]\[ R(x, y) = 140x + 180y + 0.03xy - 0.07x^2 - 0.02y^2 \][/tex]
[tex]\[ C(x, y) = 3x + 4y + 20,000 \][/tex]
Profit [tex]\( P(x, y) \)[/tex] can be expressed as:
[tex]\[ P(x, y) = R(x, y) - C(x, y) \][/tex]
To find [tex]\( P_x \)[/tex] and [tex]\( P_y \)[/tex], we start by computing the partial derivatives of [tex]\( R \)[/tex] and [tex]\( C \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
### Step-by-Step Solution:
1. Partial derivative of [tex]\( R \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ R_x = \frac{\partial R}{\partial x} = 140 + 0.03y - 0.14x \][/tex]
2. Partial derivative of [tex]\( R \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ R_y = \frac{\partial R}{\partial y} = 180 + 0.03x - 0.04y \][/tex]
3. Partial derivative of [tex]\( C \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ C_x = \frac{\partial C}{\partial x} = 3 \][/tex]
4. Partial derivative of [tex]\( C \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ C_y = \frac{\partial C}{\partial y} = 4 \][/tex]
Now, the partial derivatives of the profit function [tex]\( P \)[/tex] at the point [tex]\((x, y)=(1400, 1800)\)[/tex] can be found by substituting [tex]\( x = 1400 \)[/tex] and [tex]\( y = 1800 \)[/tex] into the expressions for [tex]\( R_x \)[/tex] and [tex]\( R_y \)[/tex]:
### Calculating [tex]\( P_x \)[/tex] at [tex]\((1400, 1800)\)[/tex]:
[tex]\[ P_x = R_x - C_x \][/tex]
Substituting [tex]\( x = 1400 \)[/tex] and [tex]\( y = 1800 \)[/tex] into [tex]\( R_x \)[/tex]:
[tex]\[ R_x = 140 + 0.03(1800) - 0.14(1400) = 140 + 54 - 196 = -2 \][/tex]
[tex]\[ C_x = 3 \][/tex]
Thus,
[tex]\[ P_x = R_x - C_x = -2 - 3 = -5 \][/tex]
So, [tex]\( P_x(1400, 1800) = -5 \)[/tex].
### Calculating [tex]\( P_y \)[/tex] at [tex]\((1400, 1800)\)[/tex]:
[tex]\[ P_y = R_y - C_y \][/tex]
Substituting [tex]\( x = 1400 \)[/tex] and [tex]\( y = 1800 \)[/tex] into [tex]\( R_y \)[/tex]:
[tex]\[ R_y = 180 + 0.03(1400) - 0.04(1800) = 180 + 42 - 72 = 150 \][/tex]
[tex]\[ C_y = 4 \][/tex]
Thus,
[tex]\[ P_y = R_y - C_y = 150 - 4 = 146 \][/tex]
So, [tex]\( P_y(1400, 1800) = 146 \)[/tex].
### Interpretation of Results:
- [tex]\( P_x(1400, 1800) = -5 \)[/tex] implies that, at the production level of 1400 type [tex]\( A \)[/tex] calculators and 1800 type [tex]\( B \)[/tex] calculators, increasing the production of type [tex]\( A \)[/tex] calculators by one unit is expected to decrease the profit by 5 units.
- [tex]\( P_y(1400, 1800) = 146 \)[/tex] implies that, at the same production level, increasing the production of type [tex]\( B \)[/tex] calculators by one unit is expected to increase the profit by 146 units.
These partial derivatives provide a measure of the sensitivity of profit to changes in the production quantities of each type of calculator.
Given the functions for revenue [tex]\( R(x, y) \)[/tex] and cost [tex]\( C(x, y) \)[/tex],
[tex]\[ R(x, y) = 140x + 180y + 0.03xy - 0.07x^2 - 0.02y^2 \][/tex]
[tex]\[ C(x, y) = 3x + 4y + 20,000 \][/tex]
Profit [tex]\( P(x, y) \)[/tex] can be expressed as:
[tex]\[ P(x, y) = R(x, y) - C(x, y) \][/tex]
To find [tex]\( P_x \)[/tex] and [tex]\( P_y \)[/tex], we start by computing the partial derivatives of [tex]\( R \)[/tex] and [tex]\( C \)[/tex] with respect to [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
### Step-by-Step Solution:
1. Partial derivative of [tex]\( R \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ R_x = \frac{\partial R}{\partial x} = 140 + 0.03y - 0.14x \][/tex]
2. Partial derivative of [tex]\( R \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ R_y = \frac{\partial R}{\partial y} = 180 + 0.03x - 0.04y \][/tex]
3. Partial derivative of [tex]\( C \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ C_x = \frac{\partial C}{\partial x} = 3 \][/tex]
4. Partial derivative of [tex]\( C \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ C_y = \frac{\partial C}{\partial y} = 4 \][/tex]
Now, the partial derivatives of the profit function [tex]\( P \)[/tex] at the point [tex]\((x, y)=(1400, 1800)\)[/tex] can be found by substituting [tex]\( x = 1400 \)[/tex] and [tex]\( y = 1800 \)[/tex] into the expressions for [tex]\( R_x \)[/tex] and [tex]\( R_y \)[/tex]:
### Calculating [tex]\( P_x \)[/tex] at [tex]\((1400, 1800)\)[/tex]:
[tex]\[ P_x = R_x - C_x \][/tex]
Substituting [tex]\( x = 1400 \)[/tex] and [tex]\( y = 1800 \)[/tex] into [tex]\( R_x \)[/tex]:
[tex]\[ R_x = 140 + 0.03(1800) - 0.14(1400) = 140 + 54 - 196 = -2 \][/tex]
[tex]\[ C_x = 3 \][/tex]
Thus,
[tex]\[ P_x = R_x - C_x = -2 - 3 = -5 \][/tex]
So, [tex]\( P_x(1400, 1800) = -5 \)[/tex].
### Calculating [tex]\( P_y \)[/tex] at [tex]\((1400, 1800)\)[/tex]:
[tex]\[ P_y = R_y - C_y \][/tex]
Substituting [tex]\( x = 1400 \)[/tex] and [tex]\( y = 1800 \)[/tex] into [tex]\( R_y \)[/tex]:
[tex]\[ R_y = 180 + 0.03(1400) - 0.04(1800) = 180 + 42 - 72 = 150 \][/tex]
[tex]\[ C_y = 4 \][/tex]
Thus,
[tex]\[ P_y = R_y - C_y = 150 - 4 = 146 \][/tex]
So, [tex]\( P_y(1400, 1800) = 146 \)[/tex].
### Interpretation of Results:
- [tex]\( P_x(1400, 1800) = -5 \)[/tex] implies that, at the production level of 1400 type [tex]\( A \)[/tex] calculators and 1800 type [tex]\( B \)[/tex] calculators, increasing the production of type [tex]\( A \)[/tex] calculators by one unit is expected to decrease the profit by 5 units.
- [tex]\( P_y(1400, 1800) = 146 \)[/tex] implies that, at the same production level, increasing the production of type [tex]\( B \)[/tex] calculators by one unit is expected to increase the profit by 146 units.
These partial derivatives provide a measure of the sensitivity of profit to changes in the production quantities of each type of calculator.
