PRELIMINARY AND INCOMPLETE - DO NOT CIRCULATE

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## 11.2 Cost Minimization

## Isocost Lines

## The cost minimization problem

$$\begin{aligned} \min_{L,K} \ \ & wL + rK\\ \text{s.t. } \ & f(L,K) = q \end{aligned}$$

## The “gravitational pull” along an isoquant

## When calculus works

$$\mathcal{L}(L,K,\lambda) = wL + rK + \lambda [q - f(L,K)]$$

which has the three first-order conditions

Setting the $\lambda$ for the first two conditions equal to each other gives us the condition

$$\frac{MP_L}{MP_K} = \frac{w}{r}$$

## Kinks and corners

## Worked example

For this chapter, we’re going to use the production function

$$\frac{K}{L} = \frac{w}{r} \Rightarrow K = \frac{w}{r}L$$

Plugging this into the constraint (i.e., the isoquant) gives us

$$K^\star = \frac{w}{r}L^\star = \sqrt \frac{w}{r} \times q$$

## 7 Minimizing Costs

The policy question will an increase in the minimum wage decrease employment.

## Exploring the Policy Question

- What potential benefits to raising the minimum wage are identified in the article?
- What potential disadvantages to raising the minimum wage are identified?

## Learning Objectives

7.1 the economic concept of cost.

## 7.2 Short-Run Cost Minimization

Learning Objective 7.2 : Describe the solution to the cost minimization problem in the short run.

## 7.3 Long-Run Cost Minimization

Learning Objective 7.3 : Describe the solution to the cost minimization problem in the long run.

## 7.4 When Input Costs and Output Change

Learning Objective 7.4 : Analyze the effect of changes in prices or output on total cost.

## 7.5 Policy Example Will an Increase in the Minimum Wage Decrease Employment?

Learning Objective 7.5 : Apply the concept of cost minimization to a minimum-wage policy.

## Fixed and Variable Costs

## Opportunity Costs

## Depreciation

In chapter 6 , we studied the short-run production function:

[latex]Q=f(L,\overline{K})[/latex],

[latex]Q=f(\overline{K},L)[/latex]

[latex]L^*=f^{−1}(\overline{K},q)[/latex]

Let’s consider a specific example of a Cobb-Douglas production function:

[latex]Q=10\overline{K}^{\frac{1}{2}}L^{\frac{1}{2}}[/latex]

[latex]L^{\frac{1}{2}}=(\frac{q}{10\overline{K}^{\frac{1}{2}}})^2[/latex]

[latex]L^*=\frac{q^2}{100\overline{K}}[/latex]

## Total Cost in the Long Run and the Isocost Line

- The cost of labor is called the wage rate, [latex]w[/latex].
- The cost of capital is called the rental rate, [latex]r[/latex].
- The cost of the labor input is the wage rate multiplied by the amount of labor employed, [latex]wL[/latex].
- The cost of capital is the rental rate multiplied by the amount of capital, [latex]rK[/latex].

The total cost ([latex]C[/latex]), therefore, is

[latex]MRTS=-\frac{MP_L}{MP_K}[/latex]

So the solution to the long-run cost minimization problem is

[latex]MRTS=-\frac{w}{r}[/latex],

[latex]\frac{MP_L}{MP_K}=\frac{w}{r}[/latex] (7.1)

This can be rearranged to help with intuition:

[latex]\frac{MP_L}{w}=\frac{MP_K}{r}[/latex] (7.2)

[latex]\frac{MP_L}{w}< \frac{MP_K}{r}[/latex]

## An Example of Minimizing Costs in the Long Run

Calculus (long-run cost minimization problem).

[latex]\frac{min}{(L,K)}{wL+rK}[/latex] (7.1C)

subject to [latex]Q=f=(L,K)[/latex] (7.2C)

We can proceed by defining a Lagrangian function:

(7.3C) [latex]\wedge (L,K,\lambda) = wL+rK-\lambda (f(L,K)-q)[/latex]

(7.6C) [latex]\frac{\partial \wedge }{\partial \lambda }=0\Rightarrow Q=f(L,K)[/latex]

(7.7C) [latex]\frac{MP_L}{MP_K}=\frac{w}{r}[/latex]

(7.8C) [latex]Q=f(L,K)[/latex]

(7.3) [latex]MP_L=5L^{-\frac{1}{2}}K^{\frac{1}{2}}[/latex]

(7.4) [latex]MP_K=5L^{\frac{1}{2}}K^{-\frac{1}{2}}[/latex]

And thus the [latex]MRTS,-\frac{MP_L}{MP_K}[/latex] is [latex]-\frac{K}{L}[/latex].

The ratio [latex]-\frac{w}{r}[/latex] in this case is [latex]-\frac{20}{5}[/latex], or −4.

(7.5) [latex]1,000=10L^{\frac{1}{2}}K^{\frac{1}{2}}[/latex]

- Find the marginal product of labor.
- Find the marginal product of capital.
- Find the MRTS.
- Find the optimal amount of labor and capital inputs.
- For [latex]Q=5L^{\frac{1}{3}}K^{\frac{2}{3}}[/latex] and [latex]w[/latex] and [latex]r[/latex], solve for the input-demand functions using the Lagrangian method.

The production function for one thousand barrels has not changed:

[latex]1,000=10L^{\frac{1}{2}}K^{\frac{1}{2}}[/latex]

[latex]C(Q) = 20(L) + 5(K) = $20(50) + $5(200) = $1,000 + $1,000 = $2,000[/latex].

Now when capital rental rates increase to $10, total cost becomes

[latex]C(Q) = 20(L) + 10(K) = $20(71) + $10(142) = $1,420 + $1,420 = $2,840[/latex].

## Expansion Path

(7.6) [latex]Q=10L^{\frac{1}{2}}K^{\frac{1}{2}}[/latex]

- If [latex]K=4L[/latex], then [latex]Q=10L^{\frac{1}{2}}(4L)^{\frac{1}{2}}[/latex], or [latex]Q=20L[/latex] or [latex]L(q)={\frac{q}{20}}[/latex].
- If [latex]L(q)=\frac{q}{20’}[/latex], then [latex]K(q)=\frac{4q}{20}=\frac{q}{5}[/latex].

- At [latex]Q=2,000[/latex]
- [latex]L(2,000)=\frac{2,000}{20}=100[/latex] and [latex]K(2,000)=\frac{2,000}{5}=400[/latex].
- At [latex]Q=3,000[/latex]
- [latex]L(3,000)=\frac{3,000}{20}=150[/latex] and [latex]K(3,000)=\frac{3,000}{5}=600[/latex].

[latex]C(Q)=wL+rK=wL(Q)+rK(Q)=w\frac{Q}{20}+r\frac{q}{5}[/latex]

Graphically, the expansion path and associated long-run total cost curve look like figure 7.4.1 .

Solving for the input demand for capital yields

[latex]K^*=\frac{Q}{10}(\frac{w}{r})^{\frac{1}{2}}[/latex].

Since [latex]L=\frac{rK}{w}[/latex], we can find the input demand for labor:

[latex]L^*=\frac{Q}{10}(\frac{r}{w})^{\frac{1}{2}}[/latex]

## Perfect Complement and Perfect Substitute Production Functions

Recall that a perfect substitute production function is of the additive type:

[latex]Q=\alpha L+\beta K[/latex]

- Do you support a national minimum wage increase? Why or why not?
- Do you think the benefits of a minimum wage increase outweigh the costs? Explain your answer.
- What do you predict would happen if, instead of a minimum wage, a tax on the purchase or rental of capital equipment was imposed?

## Review: Topics and learning outcomes

A cost that does not change as output changes.

A cost that changes as output changes.

The loss of value of a durable good or asset over time.

A good that has a long usable life.

A function that describes the optimal factor input level for every possible level of output, i.e.,

A graph of every possible combination of inputs that yields the same cost of production.

The MRTS is also the slope of the isoquant

## Long-run cost minimization problem

The slope of the isoquant is the MRTS, and the slope of the is

[latex]MRTS=-\frac{w}{r}[/latex]

[latex]\frac{MP_L}{w}=\frac{MP_K}{r}[/latex]

This formula has many different calculus derived conclusions that should be reviewed.

[latex]\frac{MP_L}{MP_K}=\frac{w}{r}[/latex]

[latex]\frac{\partial \wedge }{\partial \lambda }=0\Rightarrow Q=f(L,K)[/latex]

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## Solve a Minimization Problem Using Linear Programming

## Sign up for the Dummies Beta Program to try Dummies' newest way to learn.

- Choose variables to represent the quantities involved. The quantities here are the number of tablets. Let a tablet of Vega Vita be represented by v and a tablet of Happy Health be represented by h .
- Write an expression for the objective function using the variables. The goal is to spend the smallest amount of money necessary (so you want the minimum). Vega Vita costs 20 cents per tablet, and Happy Health costs 30 cents per tablet. Minimize: $0.20 v + $0.30 h

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Solving cost minimization problems

Situation: Goal: minimize TC = PL . L + PK . K Constraint: produce amount Qo = Q(L,K)

## How can we solve minimization problem using simplex method

## How can we solve linear programming problem using simplex method?

- Standard form.
- Introducing slack variables.
- Creating the tableau.
- Pivot variables.
- Creating a new tableau.
- Checking for optimality.
- Identify optimal values.

## What is simplex method maximization?

- Given n decision variables, usually converges in O(n) operations with O(n) pivots.
- Takes advantage of geometry of problem: visits vertices of feasible set and checks each visited vertex for optimality. (In primal simplex, the reduced cost can be used for this check.)
- Good for small problems.

## What is minimization and maximization?

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## How to solve a minimization problem of a least squares cost function?

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## Solving the following cost minimization problem using Kuhn-Tucker conditions

I set up the Lagrange in the following way:

$$\mathcal{L}=4\sqrt{y_{1}}+2\sqrt{y_{2}}-\lambda(y_{1}+y_{2}-y)-\mu_{1}y_{1}-\mu_{2}y_{2}$$

That is, using principal minors, we have:

$$\mid H \mid ^{(1)}<0, \ \ and \ \ \mid H \mid ^{(2)}=|H|>0$$

That is, $$|H|^{(1)}=det(-\frac{5\sqrt{5}}{8y^{3/2}})=-\frac{5\sqrt{5}}{8y^{3/2}}<0$$

$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix}$$ Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.

- $\begingroup$ Thank you, this was a rookie mistake. This next question is more of a math question, why am I finding that |H|^(1)<0 and |H|^(2)=|H|>0 (ie. since the upper left entry is negative and the lower right entry is negative). Wouldn't the determinant be positive? $\endgroup$ – Mistah White Oct 20, 2021 at 16:08
- $\begingroup$ What exactly are you denoting by $|H|^{(1)}<0$? $\endgroup$ – Giskard Oct 20, 2021 at 16:10
- $\begingroup$ @MistahWhite Perhaps you should edit your calculations into the body of the question, as otherwise it may be nigh-on-impossible to determine why you are reaching erroneus results. $\endgroup$ – Giskard Oct 20, 2021 at 16:11
- $\begingroup$ Thank you for your suggestion. I just edited the calculations into the question. $\endgroup$ – Mistah White Oct 20, 2021 at 16:18
- $\begingroup$ @MistahWhite I think you may have misread Sylvester's criterion ? Or why is it that you think this contradicts negative definiteness? $\endgroup$ – Giskard Oct 20, 2021 at 16:21

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Lagrangian: How to Solve Cost Minimization Problem - YouTube hey guys! so today we'll talk about the lagrangian cost minimization problem and in this picture Macroeconomics: Supply and...

The goal of the firm's cost minimization problem is to produce a given quantity at the lowest possible cost: that is, find the point along an isoquant which is along the lowest possible isocost line. The key thing here is that we're treating the amount of output as fixed; that is, some target amount q q.

In this episode I describe the Cost Minimization Problem in case of two inputs and show how we solve for optimal cost functions.Important Note for Navigating...

The short-run cost minimization problem is straightforward: since the only adjustable input is labor, the solution to the problem is to employ just enough labor to produce a given level of output. Figure 7.2.1 illustrates the solution to the short-run cost minimization problem.

Solving the cost-minimization problem Conceptually: find best (lowest) isocost given the required isoquant. Graphically... Draw the "target" output level (isoquant) q 0. Ask: what is the cheapest combination of L,K that makes q 0 possible. Solving the cost-minimization problem What if prices of factors change? What if target output changes?

The following is a minimization problem dealing with saving money on supplements. You're on a special diet and know that your daily requirement of five nutrients is 60 milligrams of vitamin C, 1,000 milligrams of calcium, 18 milligrams of iron, 20 milligrams of niacin, and 360 milligrams of magnesium.

Solving cost minimization problems Situation: Goal: minimize TC = PL. L + PK. K Constraint: produce amount Qo = Q (L,K) Key relationships: (1) Tangency Condition (tc): MPL/MPK = ( D Q/ D L)/ ( D Q/ D K) = PL/PK (2) Output Constraint: Qo = Q (L,K) To do: Try the following example: Given: Q = L 1/2 K 1/2 PL = $4, PK = $1

A fully worked example going through how to find cost-minimizing combinations of inputs with three classic production functions: linear, Leontief, and Cobb-Douglas. Regression Exercise Solution...

For those of you interested in math and knowing what the Lagrange method of solving optimization problems is, the cost minimization problem of the rm stated above can be solved also using this method. What we do is write the Lagrangean: ( x 1;x 2; ) = w 1x 1 +w 2x 2 (f(x 1;x 2) y) where is the Lagrange multiplier on the constraint.

To every minimization problem there corresponds a dual problem. The solution of the dual problem is used to find the solution of the original problem. The dual problem is a maximization problem, which we learned to solve in the last section. We first solve the dual problem by the simplex method.

This video uses a lagrangian to minimize the cost of producing a given level of output.

Solution: The given maximization problem is converted into minimization problem by subtracting from the highest sales value (i.e., 41) with all elements of the given table. Reduce the matrix column-wise and draw minimum number of lines to cover all the zeros in the matrix, as shown in Table.

The ratio pi=MUi measures the cost of increasing utility by one util, or the \cost{per{bang". At the optimum the agent equates the cost{per{bang of the two goods. Intuitively, if good 1 has a higher cost{per{bang than good 2, then the agent should spend less on good 1 and more on good 2. In doing so, she could attain the same utility at a lower ...

% Minmization optimization while (sum (abs (beta_d-beta_u))>0.1) initial_guess = randn (2,2); OLS = @ (B,input_vars)sum (abs (myfun (B,input_vars)-beta_u).^2); % ordinary least squares cost function opts = optimoptions (@fminunc, 'MaxIterations', 10000, 'MaxFunctionEvaluations', 50000, 'Display', 'Iter', 'FiniteDifferenceStepSize', 1e-3);

I can take the ratio to find that 2 y 2 y 1 = 1 Using the fact that y 1 + y 2 = y, we can see that we have the following solution y 1 = 4 5 y a n d y 2 = 1 5 y I now check the second order conditions. We have C ( y) = 4 y 1 + 2 y 2 which means C 1 = 2 y 1, C 2 = 1 y 2 C 11 = − 1 y 1 3 / 2, C 22 = − 1 2 y 2 3 / 2, C 12 = C 21 = 0

Describes how to set up and solve cost minimization problems. AboutPressCopyrightContact usCreatorsAdvertiseDevelopersTermsPrivacyPolicy & SafetyHow YouTube worksTest new features © 2022...

Cost Minimization Given factors of production and , with rental prices and , find cheapest way to produce a given level of output : such that: wK K L wL y ()wLL wK K L K +, min y = F(L,K) Cost Function Solution to minimization problem is cost function: wLL ()wL,wK , y wK K wL,wK, y * + * c()wL,wK, y = Finding the Cost Function Cost of using and

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In this video we learn how to solve cost minimization problem of a firm with production function given by f(l,k) = min{2l+k, l+2k}. And we also solve profit...

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Solving cost minimization problems. A fully worked example going through how to find cost-minimizing combinations of inputs with three classic production functions: linear ` minimum calculator. Free Minimum Calculator - find the Minimum of a data set step-by-step. Determine math question ...

Cost Minimization with a Lagrangian

Minimization linear programming problems are solved in much the same way as the maximization problems. For the standard minimization linear program, the constraints are of the form a x + b y ≥ c, as opposed to the form a x + b y ≤ c for the standard maximization problem.

Solve a Minimization Problem Using Linear Programming. ... Subsequently, we show how linear programming can be used to solve complex constrained profit maximization and cost minimization problems, and we estimate the. 24/7 Customer Help If you need help, we're here for you 24/7. ...