Padp 6950 Fall 2020 Lawler Uga Homework 5 Due October 26 ✓ Solved

Padp 6950 Fall 2020 Lawler Ugahomework 5 Due October 261 A Local Fa

Padp 6950 Fall 2020 Lawler Ugahomework 5 Due October 261 A Local Fa

Analyze the production functions and cost minimization problems faced by a farmer growing peaches, as well as the advertising campaign strategy for a political candidate, based on the specified production functions and cost parameters. Your discussion should include calculations of isoquants, short-run and long-run production functions, marginal analysis, and cost-effective advertising combinations. Use appropriate economic concepts and graphing to illustrate the solutions clearly and comprehensively.

Sample Paper For Above instruction

Part 1: Farmer's Peach Production Analysis

Problem Statement:

A farmer produces peaches using land (K) and labor (L) with an production function:

\[ Q(K, L) = K^{0.5}L^{0.5} \]

Given the output level of 6 bushels, determine input combinations, plot the isoquant, analyze short-run and long-run production, and evaluate marginal outputs under constraints.

Input Combinations for 6 Bushels of Peaches:

To find input combinations (K, L) that yield 6 bushels:

\[ K^{0.5}L^{0.5} = 6 \]

which simplifies to:

\[ \sqrt{K} \times \sqrt{L} = 6 \]

\[ \sqrt{K \times L} = 6 \]

\[ K \times L = 36 \]

Therefore, any (K, L) satisfying:

\[ L = \frac{36}{K} \]

will produce 6 bushels. For example:

- If \(K = 4\), then \(L = 36/4 = 9\).

- If \(K = 9\), then \(L = 36/9 = 4\).

- If \(K = 1\), then \(L = 36/1 = 36\).

Plotting these on a graph with land (K) on the y-axis and labor (L) on the x-axis results in a rectangular hyperbola illustrating all combinations yielding 6 bushels. This represents the isoquant curve for Q=6, indicating various trade-offs between land and labor.

Short-Run Production Function (Fixed Land):

Given that in the short run, the farmer has only 4 units of land:

\[ Q_{short}(L) = 4^{0.5} \times L^{0.5} = 2 \times L^{0.5} \]

Graphing \(Q_{short}\) for \(L\) from 0 to 16 yields:

\[ Q_{short}(L) = 2 \sqrt{L} \]

This curve is a square root function indicating diminishing marginal returns to labor.

Marginal Product of Labor (MPL):

The MPL at a given level of L when land is fixed at 4:

\[ MPL(L) = \frac{dQ_{short}}{dL} = 2 \times \frac{1}{2} L^{-0.5} = \frac{1}{\sqrt{L}} \]

- When \(L=1\), \(MPL = 1\).

- When \(L=4\), \(MPL= \frac{1}{2}\).

Thus, adding the first unit of labor (from 0 to 1) yields 1 bushel of additional output; adding from 1 to 2 units of labor yields:

\[ \frac{1}{\sqrt{2}} \approx 0.707 \] bushels.

Additional Output for Different Labor Inputs:

- From 1 to 2 units of labor:

\[ \Delta Q \approx 0.707 \text{ bushels} \]

- From 4 to 5 units:

\[ \Delta Q \approx \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}} \approx 0.5 - 0.447 = 0.053 \text{ bushels} \]

indicating diminishing marginal returns.

Long-Run Adjustment (Larger Land):

When the land size increases to 16 units:

\[ Q_{long}(L) = 16^{0.5} \times L^{0.5} = 4 \times \sqrt{L} \]

The graph of output as a function of labor when land is fixed at 16 shifts the curve upward, showing increased capacity.

Graphically, both the short-run and long-run production functions are plotted with L on the x-axis and output on the y-axis, illustrating the impact of fixed versus variable land.

Part 2: Political Advertising Campaign Optimization

Given Data:

Production function for votes:

\[ V(T, NP) = 300 \times T^{0.6} \times NP^{0.2}, \]

costs:

- \(C_T = \$800\) per TV ad

- \(C_{NP} = \$200\) per newspaper ad

Target:

\[ V(T, NP) \geq 1800 \text{ votes} \]

Marginal products:

\[ MP_{T} = \frac{\partial V}{\partial T} = 180 \times T^{-0.4} \times NP^{0.2} \]

\[ MP_{NP} = \frac{\partial V}{\partial NP} = 60 \times T^{0.6} \times NP^{-0.8} \]

Optimization for Cost-Effective Advertising:

To minimize total cost:

\[ C_{total} = 800T + 200NP \]

subject to:

\[ 300 T^{0.6} NP^{0.2} \geq 1800. \]

Using the method of Lagrange multipliers:

\[ \mathcal{L} = 800T + 200NP - \lambda ( 300 T^{0.6} NP^{0.2} - 1800). \]

Set derivatives to zero:

\[

\frac{\partial \mathcal{L}}{\partial T} = 800 - \lambda \times 300 \times 0.6 T^{-0.4} NP^{0.2} = 0,

\]

\[

\frac{\partial \mathcal{L}}{\partial NP} = 200 - \lambda \times 300 T^{0.6} \times 0.2 NP^{-0.8} = 0,

\]

\[

\frac{\partial \mathcal{L}}{\partial \lambda} = 300 T^{0.6} NP^{0.2} - 1800 = 0.

\]

Dividing the first EOS by the second:

\[

\frac{800}{200} = \frac{0.6 T^{-0.4} NP^{0.2}}{0.2 T^{0.6} NP^{-0.8}} \Rightarrow 4 = 3 \times T^{-1} \times NP.

\]

Rearranged:

\[

NP = \frac{4}{3} T.

\]

From the production constraint:

\[

300 T^{0.6} (NP)^{0.2} = 1800,

\]

substitute \(NP = \frac{4}{3} T\):

\[

300 T^{0.6} \left(\frac{4}{3} T\right)^{0.2} = 1800,

\]

\[

300 T^{0.6} \times \left(\frac{4}{3}\right)^{0.2} T^{0.2} = 1800,

\]

\[

300 \times \left(\frac{4}{3}\right)^{0.2} T^{0.8} = 1800,

\]

\[

T^{0.8} = \frac{1800}{300 \times (4/3)^{0.2}}.

\]

Calculate \((4/3)^{0.2}\):

\[

(4/3)^{0.2} \approx e^{0.2 \ln(4/3)} \approx e^{0.2 \times 0.2877} \approx e^{0.0575} \approx 1.059.

\]

Thus:

\[

T^{0.8} \approx \frac{1800}{300 \times 1.059} = \frac{1800}{317.7} \approx 5.668,

\]

\[

T = (5.668)^{1/0.8} \approx (5.668)^{1.25} = e^{1.25 \times \ln 5.668} \approx e^{1.25 \times 1.735} = e^{2.169} \approx 8.76.

\]

Correspondingly:

\[

NP = \frac{4}{3} \times 8.76 \approx 11.68.

\]

Calculate total cost:

\[

C_{total} = 800 \times 8.76 + 200 \times 11.68 \approx 7008 + 2336 = \$9,344.

\]

Therefore, the optimal combination involves approximately 8.76 TV ads and 11.68 newspaper ads, costing about \$9,344 in total.

Summary and Conclusion

The farmer’s crop production illustrates the application of isoquants, short-run, and long-run analysis, highlighting diminishing marginal returns. Meanwhile, the political advertising problem demonstrates cost minimization under a production constraint, employing calculus and Lagrangian optimization to find the lowest-cost strategy for reaching the vote target efficiently.

References

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