Question

Please help me i am preparing for my final exam but i don't know how to answer the question like this.

Given the estimated production function of rice in Kelantan by using two inputs; nitrogen (Xl) and phosphate (X2) are as below:

Q = 200 + 4X1 + 77X2 - 0.05X1^2 - o.5X2^2+ 0.10X1X2


The other information regarding the output price of rice is equivalent to RMI per kg, the cost per unit of nitrogen is RM4 per kg, phosphate per unit cost is RM5 per kg and the total money with the farmer is RM700.


Calculate the optimum level of the use of nitrogen and phosphate input.



Calculate the optimum output.





Given the Cobb-Douglas production function is Q = 30X^0.5The total fixed cost is RM500, the input price is RM25 per unit and the output price is RM8 per unit.

Calculate the optimum input.

Calculate the profit/loss obtained if the optimum input level has been used in the production.

Given the present value for timber is RM5000. What is the value of timber in the 29 year when the interest rate is 5 %?

Answer 1

Hey! Don't worry. I know you'll pass and I believe in you.

1. To calculate the optimum level of nitrogen and phosphate inputs, we need to maximize the production function Q = 200 + 4X1 + 77X2 - 0.05X1^2 - 0.5X2^2 + 0.10X1X2.

2. Given that the cost per unit of nitrogen is RM4 per kg and phosphate is RM5 per kg, and the total money with the farmer is RM700, we can set up the cost constraint: 4X1 + 5X2 ≤ 700.

3. To find the optimum levels, we need to take partial derivatives of the production function with respect to X1 and X2.

4. Taking the partial derivative of Q with respect to X1, we get: dQ/dX1 = 4 - 0.1X1 + 0.10X2. Taking the partial derivative of Q with respect to X2, we get: dQ/dX2 = 77 - X2 + 0.10X1.

5. Setting the partial derivatives equal to zero, we have: 4 - 0.1X1 + 0.10X2 = 0 and 77 - X2 + 0.10X1 = 0.

6. Solving these equations will give us the values of X1 and X2 that maximize the production function.

7. After obtaining the critical points, we need to check the second partial derivatives to ensure they correspond to a maximum. If the second partial derivatives are negative, then the critical points represent a maximum.

8. To calculate the optimum output, substitute the values of X1 and X2 into the production function Q = 200 + 4X1 + 77X2 - 0.05X1^2 - 0.5X2^2 + 0.10X1X2.

To calculate the profit/loss, we need to consider the total fixed cost and the output price. Given the Cobb-Douglas production function Q = 30X^0.5, the total fixed cost is RM500, and the output price is RM8 per unit.

1. To calculate the optimum input level, we need to maximize the production function Q = 30X^0.5 while considering the cost constraint.

2. The cost constraint can be set up as: input price * input level ≤ total fixed cost.

3. Substitute the values into the production function and the cost constraint equation to find the optimum input level.

To calculate the profit/loss, subtract the total cost from the total revenue. Total revenue is obtained by multiplying the output level with the output price.

Given the present value for timber is RM5000 and an interest rate of 5%, we need to find the value of timber in the 29th year.

To calculate the future value of timber, we can use the formula: Future Value = Present Value * (1 + Interest Rate)^Number of Years.

Substitute the values into the formula to find the value of timber in the 29th year.