(Transcript of audio with descriptions. Transcript includes narrator headings and description headings of the visual content)
(Speaker)
In this problem, we are asked to calculate the opportunity cost of increasing potato production in 200 pound increments. The easiest way to find the opportunity cost is to calculate the slope between each output combination. We will start our basic graph, which has points A through А labeled.
We will start at point A and calculate the slope between points A and B. The easiest way to calculate the slope is by taking the rise divided by the run.
(Description)
A vector from point, A, with coordinates, 1000 and 0, to a point with coordinates, 1000 and 300, is drawn. Another vector from a point with coordinates, 1000 and 300, to point B, with coordinates, 800 and 300, is drawn. These two vectors make a right angle.
(Speaker)
For this problem, the rise is the increase in fish production and the run is the decrease in potato production. From point A to point B, we gained 300 pounds of fish,
(Description)
Text +300 fish is typed next to the vector which has the starting point with coordinates, 1000 and 0, and the ending point with coordinates, 1000 and 300.
(Speaker)
but must give up 200 pounds of potatoes.
(Description)
Text -200 potatoes is typed next to the vector which has the starting point with coordinates, 1000 and 300, and the ending point with coordinates, 800 and 300.
(Speaker)
We can connect points A and B with a straight line and label this point negative 300 divided by 200 where the negative sign represents the reduction in potato output to increase fish output.
(Description)
The points, A, and, B, are connected with a straight line, and the line is labeled as negative 300 divided by 200.
(Speaker)
We will next calculate the slope between point B and C.
(Description)
A vector from point, B, with coordinates, 800 and 300, to a point with coordinates, 800 and 500, is drawn. Another vector from point with coordinates, 800 and 500, to point C, with coordinates, 600 and 500, is drawn. These two vectors make a right angle.
(Speaker)
Moving from B to C shows an increase in fish output of 200 pounds,
(Description)
Text +200 fish is typed next to the vector which has the starting point with coordinates, 800 and 300, and the ending point with coordinates, 800 and 500.
(Speaker)
but we must give up 200 pounds of potato production.
(Description)
Text -200 potatoes is typed next to the vector which has the starting point with coordinates, 800 and 500, and the ending point, C, with coordinates, 600 and 500.
(Speaker)
We can connect points B and C with a straight line and label this line negative 200 divided by 200.
(Description)
The points, B, and, C, are connected with a straight line, and the line is labeled as negative 200 divided by 200.
(Speaker)
You will notice this segment represents a one-to-one trade-off. For every pound of fish we produce, we must give up one pound of potatoes.
Continuing on, moving from C to D provide the economy with 100 pounds of fish,
(Description)
A vector from point, C, with coordinates, 600 and 500, to a point with coordinates, 600 and 600, is drawn. Text +100 fish is typed next to this vector.
(Speaker)
but we must give up 200 pounds of potato production.
(Description)
A vector from a point with coordinates, 600 and 600, to point, D, with coordinates, 400 and 600, is drawn. Text -200 potatoes is typed next to this vector.
(Speaker)
We can connect points C and D and label and label the slope of this line negative 100 divided by 200.
(Description)
The points, C, and, D, are connected with a straight line, and the line is labeled as negative 100 divided by 200.
(Speaker)
Moving from D to E, provides the economy with 50 pounds of fish,
(Description)
A vector from point, D, with coordinates, 400 and 600, to a point with coordinates, 400 and 650, is drawn. Text +50 fish is typed next to this vector.
(Speaker)
but requires giving up another 200 pounds of potatoes.
(Description)
A vector from a point with coordinates, 400 and 650, to point, E, with coordinates, 200 and 650, is drawn. Text -200 potatoes is typed next to this vector.
(Speaker)
By this point, you should notice that every 200 pound decrease in potato production results in fewer and fewer pounds of fish. Next we connect the points D and E and label the slope at negative 50 divided by 200.
(Description)
The points, D, and, E, are connected with a straight line, and the line is labeled as negative 50 divided by 200.
(Speaker)
Finally, moving from E to F results in gaining 25 pounds of fish,
(Description)
A vector from point, E, with coordinates, 200 and 650, to a point with coordinates, 200 and 675, is drawn. Text +25 fish is typed next to this vector.
(Speaker)
but must give up the last 200 pounds of potatoes.
(Description)
A vector from a point with coordinates, 200 and 675, to point, F, with coordinates, 0 and 675, is drawn. Text -200 potatoes is typed next to this vector.
(Speaker)
We connect the last two points and label the slope of this line negative 25 divided by 200.
(Description)
The points, E, and, F, are connected with a straight line, and the line is labeled as negative 25 divided by 200.
(Speaker)
Now that we have calculated the slope between each point, we can easily find the opportunity cost of increasing potato output by 200 pounds. Starting from point F, were Atlantis produced zero pounds of potatoes, we can use the slope to find the opportunity cost of producing the first 200 pounds of potatoes. As production shifts from point F to point E, Atlantis gains 200 pounds of potatoes, but gives up producing 25 pounds of fish. We can continue to find the opportunity cost of producing more and more potatoes by working down the production possibility frontier. In order for Atlantis to produce the next 200 pounds of potatoes, they must give up an additional 50 pounds of fish. This will move the economy from point E to point D. By now you should notice that the opportunity cost of increasing potatoes output by 200 pounds is the numerator of the slope between the two points.
It will also help to calculate the opportunity cost of potatoes per pound. To do this, we simply divide the opportunity costs found previously by 200. This will give us the opportunity of producing one pound of potatoes. You will also notice this is the slope of the production possibilities frontier.
(Description)
The table consists of 3 columns: Maximum annual output options, Quantity of potatoes (pounds), Opportunity cost of potatoes in terms of fish, Opportunity cost of potatoes per pound. The table consists of 6 rows.
The first row: Maximum annual output option is labeled, F, Quantity of potatoes is, 0 pounds, Opportunity cost of potatoes in terms of fish is, dash, Opportunity cost of potatoes per pound is, dash.
The second row: Maximum annual output option is labeled, E, Quantity of potatoes is, 200 pounds, Opportunity cost of potatoes in terms of fish is, 675 minus 650 equals 25, Opportunity cost of potatoes per pound is, equals 25 divided by 200 equals 1 divided by 8.
The third row: Maximum annual output option is labeled, D, Quantity of potatoes is, 400 pounds, Opportunity cost of potatoes in terms of fish is, 650 minus 600 equals 50, Opportunity cost of potatoes per pound is, equals 50 divided by 200 equals 1 divided by 4.
The fourth row: Maximum annual output option is labeled, C, Quantity of potatoes is, 600 pounds, Opportunity cost of potatoes in terms of fish is, 600 minus 500 equals 100, Opportunity cost of potatoes per pound is, equals 100 divided by 200 equals 1 divided by 2.
The fifth row: Maximum annual output option is labeled, B, Quantity of potatoes is, 800 pounds, Opportunity cost of potatoes in terms of fish is, 500 minus 300 equals 200, Opportunity cost of potatoes per pound is, equals 200 divided by 200 equals 1.
The sixth row: Maximum annual output option is labeled, A, Quantity of potatoes is, 1000 pounds, Opportunity cost of potatoes in terms of fish is, 300 minus 0 equals 300, Opportunity cost of potatoes per pound is, equals 300 divided by 200 equals 3 divided by 2.