Question

In: Economics

x=0 x=2.5 x=5 x=7.5 x=10 x=12.5 pA tAx R(x) Land Rent(x) = Revenue(x)/unit area - Transport...

x=0 x=2.5 x=5 x=7.5 x=10 x=12.5
pA
tAx
R(x)

Land Rent(x) = Revenue(x)/unit area - Transport Costs(x)/unit area or R(x) = pA - tAx for all farmed locations. Assume that in the base case p = 10, A = 100, and t = 2.

Use the table to calculate R(x)

a) Draw rent as a function of distance from the central railyard.

b) What is the land rent at x = 12.5? Explain.

c) Suppose that the price of apples doubles, so that p = 20. Draw rent as a function of distance from the central railyard. Provide an economic explanation of why the slope of the rent function is 200.

d) (2) Return to the situation where p = 10 and t = 2. Now suppose that there is an increase in agricultural productivity, which causes A to increase to 200. Draw rent as a function of distance from the central railyard.

Solutions

Expert Solution

It is given that Land Rent (x) = Revenue(x)/unit area - Transport Costs(x)/unit area

That is, R(x) = pA – tAx. It is also given that p=10, A=100, t= 2.

Therefore, the function becomes:

R(x) = (10*100) – (2*100)x

Or R(x) = 1000 – 200x

The different values of Rent have been calculated on the basis of the above function and have been summarized in the table below:

x=0

x=2.5

x=5

x=7.5

x=10

x=12.5

pA

1000

1000

1000

1000

1000

1000

tAx

0

500

1000

1500

2000

2500

R(x)

1000

500

0

(-500)

(-1000)

(-1500)

a) The rent R(x) as a function of distance (x) has been drawn in blue in fig 1. It is seen that Rent R(x) and Distance are inversely related and as the distance from the central railyard increases, the rent falls. When the farm is located in the central railyard (at x=0), the rent is the highest and when the firm is located at the farthest from the central railyard, the rent is the lowest.

Since Rent depends on distance, R(x) is measured in the vertical axis and x in the horizontal axis.

b) The land rent at x=12.5 is a negative of 1500. At point x=12.5, the farm is located farthest from the central railyard. This maximizes the transportation costs from the farm top the market, located near the central area. Because total revenue is constant and the price of apples and number of apples sold are same for all, at point x=12.5, transportation costs exceeds the total revenue by 1500.

c) When the price of apple doubles and p= 20, the rent function becomes

R(x) = (20*100) – (2*100)x

Or R(x) = 2000 – 200x.

The table below provides the new values of Rent:

x=0

x=2.5

x=5

x=7.5

x=10

x=12.5

pA

2000

2000

2000

2000

2000

2000

tAx

0

500

1000

1500

2000

2500

R(x)

2000

1500

1000

5000

0

(-500)

Here too, the slope is negative, but the new pink curve starts from a higher level (2000) and lies above the blue curve in Fig 1. Its origin increase by 1000 units (previously R(x) started from 1000, now R(x) starts from 2000), but its slope remains the same, which is why there is a parallel upward shift when p increases from 10 to 20.

Although the rent function’s slope is negative, the absolute value of the slope of the curve is 200. This is because, the slope of a function y= f(x) is given by dy/dx. From the above function, slope of R(x) = dR(x)/dx = -200. The minus sign denotes the downward slope, but the actual value of the slope is 200.

This value is also equal to the marginal rent which refers to the change in total rent due to transporting the product by 1 additional unit of distance. Here, increasing the transported distance by 1 unit, reduces rent by an absolute value of 200. Eg when x=0, R(x) = 2000. But when x=1 (one additional unit of distance transported), R(x) = 2000-200 = 1800. The difference in rent is 2000-1800 = 200.

Similarly, when x rises to 2, R(x) = 2000-200*2 = 1600. Again by transporting the apples one extra unit of distance (from x=1 to x=2), the marginal rent reduces by 1800-1600 = 200.

This is the economic interpretation of why the slope of the rent function has an absolute value of 200.

d) When p=10 and t=2, and an increase in agricultural productivity raises A to 200, the rent function becomes:

R(x) = (10*200) – (2*200)x

Or R(x) = 2000- 400x

The table below provides the new values of Rent:

x=0

x=2.5

x=5

x=7.5

x=10

x=12.5

pA

2000

2000

2000

2000

2000

2000

tAx

0

1000

2000

3000

4000

5000

R(x)

2000

1000

0

(-1000)

(-2000)

(-3000)

Rent as a function of distance is now drawn in green. It is still downward sloping and starts from origin 2000, but is steeper. With increase in distance, the rent falls faster and to a greater value. The slope of this rent function with increased productivity is greater than the slope of the previous two rent functions.


Related Solutions

The revenue of a notebook salesman is R(x)=0.01x^2+15x+100. X is measured in 10 notebooks and R(x)...
The revenue of a notebook salesman is R(x)=0.01x^2+15x+100. X is measured in 10 notebooks and R(x) is measured in hundreds of dollars. Th cost is measured   C(x)=0.02x^2+5x Calculate the profit function Calculate the marginal revenue function Calculate the marginal cost function Calculate the average cost function Calculate the marginal average cost function Calculate the marginal profit function
Find the total area between y=25-x2 and the x-axis for 0≤x≤7.5. Round your answer to three...
Find the total area between y=25-x2 and the x-axis for 0≤x≤7.5. Round your answer to three decimal places. The total area between y=25-x2 and the x-axis is what?
cm(x) N(y) x^2 xy 5 2.5 25 12.5 5.5 2 30.25 11 6 2.1 36 12.6...
cm(x) N(y) x^2 xy 5 2.5 25 12.5 5.5 2 30.25 11 6 2.1 36 12.6 6 1.8 36 10.8 7 1.5 49 10.5 Consider the following small data set of 5 pairs. This table lists the length of a small metal machine piece in cm and the corresponding force in N required to break it. It also contains some calculated values for your convenience. (a) Calculate the correlation coefficient for this set. What would you conclude about the relationship...
5). Let f : [a,b] to R be bounded and f(x) > a > 0, for...
5). Let f : [a,b] to R be bounded and f(x) > a > 0, for all x in [a,b]. Show that if f is Riemann integrable on [a,b] then 1/f : [a,b] to R, (1/f) (x) = 1/f(x) is also Riemann integrable on [a,b].
For the following exercises, the revenue generated by selling x items is given by R(x) = 2x2 + 10x. Find R′(10) and interpret.
For the following exercises, the revenue generated by selling x items is given by R(x) = 2x2 + 10x.Find R′(10) and interpret.
Estimate the area under the graph of f(x)=25−x^2 from x=0 to x=5 using 5 approximating rectangles...
Estimate the area under the graph of f(x)=25−x^2 from x=0 to x=5 using 5 approximating rectangles and right endpoints. (B) Repeat part (A) using left endpoints. (C) Repeat part (A) using midpoints.
Estimate the area under the graph of f(x) =5 sqrt x. from x=0 to x=4 using...
Estimate the area under the graph of f(x) =5 sqrt x. from x=0 to x=4 using four approximating rectangles and right endpoints. sketch the graph and rectangles. is your estimate an underestimate or overestimate? Repeat using left endpoints
Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5,...
Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5, and Titrate with 0.100 mol HCl, calculate PH at 5 mL HCl, 10 mL , 15 mL, 20 mL, 24.5mL, 25mL, 25.5mL, 30mL 40.0 mL, abd 50 mL HCl. Graph the solution with hoizontal axis being the PH up to 14 and The y-axis being mL HCl.
4. (30) Water (r = 1000 kg/m3 and µ = 1.12 x 10-3 Pa-s) is siphoned...
4. (30) Water (r = 1000 kg/m3 and µ = 1.12 x 10-3 Pa-s) is siphoned through a 2mm diameter tube 1m long, with the tube exit 0.5m below the surface of the water. a) Write the governing equation Simplify to get an equation in V only Solve for V Is the flow laminar or turbulent?
A chicken house (20' x 10'x 8') with walls and roof with R-value = 5 hr...
A chicken house (20' x 10'x 8') with walls and roof with R-value = 5 hr oF ft2/BTU. There are 2 wooden doors ( 80” x 36”) with an R-value of 2.5 hr ºF sq.ft/BTU. The inside temperature is held at 35oC and the exterior temperature is 20oF. There are 19 chickens in the house with a body temperature of 42oC that can be assumed to radiate as a sphere with a 14" diameter and a emissivity of 0.85. PART...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT