Can Mary Minimize Her Costs and Make a Profit?

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In summary, the production function for Mary Jane's herb growing business is Q = 3(K*L)^0.5, with isoquants having a slope of -L/K. The rental cost of capital is 0.20 and the opportunity cost of labor is $7.20 per hour. Mary's goal is to obtain a value added of $18,000 as cheaply as possible. By setting the derivative of the isoquant equal to the relative price of labor and capital, we can find the optimal values for K and L that will minimize costs and allow Mary to make a profit.
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kristo
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Homework Statement



Mary Jane grows herbs in her attic. The relationship between her inputs and outputs is given by the production function is Q = 3(K*L)^0.5 , where Q is value added (value of gross output less cost of electricity, fertilizer, seed, and food for a guard dog), K is the value of capital goods (lamps, sprinklers, pots, and security systems), and L is Mary’s hours of labor. The isoquants corresponding to this production function have slope -L/K . The rental cost of capital (interest rate plus depreciation rate) is 0.20. The opportunity cost of Mary’s labor is $7.20 per hour. Mary’s goal is to obtain a value added of $18,000 as cheaply as possible. a) Find the optimal values of K and L. b) Can Mary make a profit?


Homework Equations





The Attempt at a Solution


First of all, I don't understand why the function's derivative is -L/K. I know that to minimize the cost I must solve the equation -L/K = Relative Price of L & K. I think relative price of L & K is (Cost of Labour Per Hour)/(Cost of Rent Per Hour), but I don't know the price of Rent Per Hour. And if anyone could explain me the reason why cost is minimized when the isoquant's derivative equals the relative price of inputs, I'd be grateful.
 
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kristo said:

The Attempt at a Solution


First of all, I don't understand why the function's derivative is -L/K.

The isoquants are curves of constant Q. Along these curves, the variation in Q is zero, so

[tex] 0 = dQ = 3 \left( \sqrt{\frac{L}{K}} dK + \sqrt{\frac{K}{L}} dL \right).[/tex]

This means that at constant Q,

[tex] \frac{dL}{dK} = - \frac{L}{K} .[/tex]

I know that to minimize the cost I must solve the equation -L/K = Relative Price of L & K. I think relative price of L & K is (Cost of Labour Per Hour)/(Cost of Rent Per Hour), but I don't know the price of Rent Per Hour. And if anyone could explain me the reason why cost is minimized when the isoquant's derivative equals the relative price of inputs, I'd be grateful.

Our cost of capital is [tex]R=0.20[/tex] while our opportunity cost of labor is [tex]C= $7.20[/tex]/hr. Our total cost is therefore [tex]T = R K + CL[/tex]. We want to minimize this subject to the condition [tex]Q= $18k[/tex]. This means that we want to minimize the function. You should find the minimum yourself, it's not quite what you wrote.
 

Related to Can Mary Minimize Her Costs and Make a Profit?

1. How is cost minimized using calculus?

Cost can be minimized using calculus by finding the derivative of the cost function and setting it equal to zero. This will give the critical points, which can then be used to determine the minimum cost.

2. What is a cost function?

A cost function is a mathematical representation of the total cost of production, taking into account factors such as labor, materials, and overhead expenses.

3. How does the marginal cost relate to calculus?

The marginal cost is the derivative of the cost function. It represents the change in cost for each additional unit produced and can be used to find the optimal production level.

4. What is the difference between average cost and marginal cost?

The average cost is the total cost divided by the number of units produced, while marginal cost is the cost of producing one additional unit. Average cost gives an overall view of production efficiency, while marginal cost helps determine the optimal production level.

5. Can calculus be used to minimize costs in all industries?

Yes, calculus can be applied to minimize costs in most industries. However, the complexity of the cost function may vary depending on the industry, and additional factors may need to be considered for a more accurate analysis.

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