Answer :
Sure, let's solve this step-by-step.
Firstly, let's understand the given functions:
- Revenue function:
[tex]\[ R(x, y) = 210x + 110y + 0.05xy - 0.07x^2 - 0.03y^2 \][/tex]
- Cost function:
[tex]\[ C(x, y) = 9x + 3y + 30,000 \][/tex]
To find the partial derivatives of these functions, we follow these steps:
1. Calculate the partial derivative of [tex]\( R(x, y) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{\partial R}{\partial x} = 210 + 0.05y - 0.14x \][/tex]
2. Calculate the partial derivative of [tex]\( R(x, y) \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ \frac{\partial R}{\partial y} = 110 + 0.05x - 0.06y \][/tex]
3. Calculate the partial derivative of [tex]\( C(x, y) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{\partial C}{\partial x} = 9 \][/tex]
4. Calculate the partial derivative of [tex]\( C(x, y) \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ \frac{\partial C}{\partial y} = 3 \][/tex]
Now let's calculate the partial derivatives of profit function [tex]\( P(x,y) \)[/tex]:
[tex]\[ P_x = \frac{\partial R}{\partial x} - \frac{\partial C}{\partial x} \][/tex]
[tex]\[ P_y = \frac{\partial R}{\partial y} - \frac{\partial C}{\partial y} \][/tex]
Let's substitute [tex]\( x = 2100 \)[/tex] and [tex]\( y = 1100 \)[/tex] into the partial derivatives.
5. Substitute [tex]\( y = 1100 \)[/tex] into [tex]\( \frac{\partial R}{\partial x} \': \[ \frac{\partial R}{\partial x} = 210 + 0.05(1100) - 0.14(2100) \] \[ \frac{\partial R}{\partial x} = 210 + 55 - 294 = -29 \] 6. Substitute \( x = 2100 \)[/tex] into [tex]\( \frac{\partial R}{\partial y} \)[/tex]:
[tex]\[ \frac{\partial R}{\partial y} = 110 + 0.05(2100) - 0.06(1100) \][/tex]
[tex]\[ \frac{\partial R}{\partial y} = 110 + 105 - 66 = 149 \][/tex]
Now calculate [tex]\( P_x \)[/tex] and [tex]\( P_y \)[/tex]:
7. Calculate [tex]\( P_x \)[/tex]:
[tex]\[ P_x = \frac{\partial R}{\partial x} - \frac{\partial C}{\partial x} = -29 - 9 = -38 \][/tex]
8. Calculate [tex]\( P_y \)[/tex]:
[tex]\[ P_y = \frac{\partial R}{\partial y} - \frac{\partial C}{\partial y} = 149 - 3 = 146 \][/tex]
Thus, the values are:
[tex]\[ P_x(2100,1100) = -38 \][/tex]
[tex]\[ P_y(2100,1100) = 146 \][/tex]
Interpretation of the Results:
- [tex]\( P_x(2100,1100) = -38 \)[/tex]: The negative value of [tex]\( -38 \)[/tex] indicates that, at the production levels of 2100 type A calculators and 1100 type B calculators, increasing the production of type A calculators by one unit would decrease the profit by 38 units.
- [tex]\( P_y(2100,1100) = 146 \)[/tex]: The positive value of [tex]\( 146 \)[/tex] indicates that, at the production levels of 2100 type A calculators and 1100 type B calculators, increasing the production of type B calculators by one unit would increase the profit by 146 units.
Firstly, let's understand the given functions:
- Revenue function:
[tex]\[ R(x, y) = 210x + 110y + 0.05xy - 0.07x^2 - 0.03y^2 \][/tex]
- Cost function:
[tex]\[ C(x, y) = 9x + 3y + 30,000 \][/tex]
To find the partial derivatives of these functions, we follow these steps:
1. Calculate the partial derivative of [tex]\( R(x, y) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{\partial R}{\partial x} = 210 + 0.05y - 0.14x \][/tex]
2. Calculate the partial derivative of [tex]\( R(x, y) \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ \frac{\partial R}{\partial y} = 110 + 0.05x - 0.06y \][/tex]
3. Calculate the partial derivative of [tex]\( C(x, y) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{\partial C}{\partial x} = 9 \][/tex]
4. Calculate the partial derivative of [tex]\( C(x, y) \)[/tex] with respect to [tex]\( y \)[/tex]:
[tex]\[ \frac{\partial C}{\partial y} = 3 \][/tex]
Now let's calculate the partial derivatives of profit function [tex]\( P(x,y) \)[/tex]:
[tex]\[ P_x = \frac{\partial R}{\partial x} - \frac{\partial C}{\partial x} \][/tex]
[tex]\[ P_y = \frac{\partial R}{\partial y} - \frac{\partial C}{\partial y} \][/tex]
Let's substitute [tex]\( x = 2100 \)[/tex] and [tex]\( y = 1100 \)[/tex] into the partial derivatives.
5. Substitute [tex]\( y = 1100 \)[/tex] into [tex]\( \frac{\partial R}{\partial x} \': \[ \frac{\partial R}{\partial x} = 210 + 0.05(1100) - 0.14(2100) \] \[ \frac{\partial R}{\partial x} = 210 + 55 - 294 = -29 \] 6. Substitute \( x = 2100 \)[/tex] into [tex]\( \frac{\partial R}{\partial y} \)[/tex]:
[tex]\[ \frac{\partial R}{\partial y} = 110 + 0.05(2100) - 0.06(1100) \][/tex]
[tex]\[ \frac{\partial R}{\partial y} = 110 + 105 - 66 = 149 \][/tex]
Now calculate [tex]\( P_x \)[/tex] and [tex]\( P_y \)[/tex]:
7. Calculate [tex]\( P_x \)[/tex]:
[tex]\[ P_x = \frac{\partial R}{\partial x} - \frac{\partial C}{\partial x} = -29 - 9 = -38 \][/tex]
8. Calculate [tex]\( P_y \)[/tex]:
[tex]\[ P_y = \frac{\partial R}{\partial y} - \frac{\partial C}{\partial y} = 149 - 3 = 146 \][/tex]
Thus, the values are:
[tex]\[ P_x(2100,1100) = -38 \][/tex]
[tex]\[ P_y(2100,1100) = 146 \][/tex]
Interpretation of the Results:
- [tex]\( P_x(2100,1100) = -38 \)[/tex]: The negative value of [tex]\( -38 \)[/tex] indicates that, at the production levels of 2100 type A calculators and 1100 type B calculators, increasing the production of type A calculators by one unit would decrease the profit by 38 units.
- [tex]\( P_y(2100,1100) = 146 \)[/tex]: The positive value of [tex]\( 146 \)[/tex] indicates that, at the production levels of 2100 type A calculators and 1100 type B calculators, increasing the production of type B calculators by one unit would increase the profit by 146 units.
