Problem 1
There are 10 bakers working at the candy store. They are real professionals and equally good at
baking cakes and muffins. In one working day each baker can bake 8 cakes or 32 muffins.
1. Draw the production possibility frontier.
2. What is the shape of the PPF?
3. One oven is broken in the bakery and as a result each baker is only able to bake 4 cakes
a day. How has the PPF changed?
4. The master came to repair the old oven and recommended the director to buy a new
oven. After the purchase was made, each baker was able to bake 16 cakes a day instead
of 8. Plot a new PPF.
Solution
1. There are 10 bakers in the confectionery, each baking either 8 cakes or 32 muffins,
respectively, the possibilities of the confectionery are either 80 cakes or 320 muffins.
2. PPF is a downward sloping line.
3. There are 10 bakers in the confectionery, each baking either 4 cakes or 32 muffins,
respectively, the possibilities of the confectionery are either 40 cakes or 320 muffins.
4.
Problem 2
Grandma sent Little Red Riding Hood into the forest to fetch berries to bake pies for the
lumberjacks.
Grandma knows that Little Red Riding Hood can collect no more than 8 kilograms of
blueberries per day, and the speed of collecting strawberries is exactly one and a half times
higher. The girl picks berries at a constant speed, and the number of berries of any kind she
picks can be any (for simplicity, we will assume that not only the number of kilograms, but
even the number of berries can be non-whole).
Grandma is going to bake a diet cake according to an old secret recipe, thanks to which the
mass of the finished cake is strictly equal to the sum of the masses of its ingredients.
The recipe says: “In each pie weighing one kilogram, you need to put 1/2 kilogram of
strawberries, 1/3 kilogram of blueberries and add the required amount of dough by weight.
There is no need to add ingredients other than dough, strawberries and blueberries to the cake.
Make a cake and place in the oven. Bake until cooked. "
The grandmother calculated exactly how many berries to bring her granddaughter and told her
to work all day at maximum productivity.
1. How many strawberries and blueberries did Little Red Riding Hood collect?
2. How many kilogram pies can grandma bake?
3. How much dough will your grandmother spend on making pies?
Solution
1 kg of blueberries can be collected in 1/8 of a day,
1 kg of strawberries in 1/12 of a day (1/8) / (1.5) = 1/12 (it can simply be indicated that 1/12 is
one and a half times less than 1/8).
For one pie you need: 1/2 kg of strawberries and 1/3 kg of blueberries.
The time spent to collect the ingredients:
1/2 * 1/12 = 1/24 days for picking strawberries
1/3 * 1/8 = 1/24 days to pick blueberries.
Then the total time to collect ingredients for one cake will be 2/24 = 1/12 days.
Then Little Red Riding Hood can collect the ingredients for 12 pies, so Grandma will bake ​12
pies.
Therefore, Little Red Riding Hood will collect ​1/2⋅12 = 6 kilograms​ of strawberries and
1/3⋅12 = 8 kilograms of blueberries​.
The cake consists of 3 parts and weighs 1 kilogram.
Strawberries and blueberries weigh 1/2 + 1/3 = 5/6 kilograms.
Since it is said that "the mass of the finished cake is strictly equal to the sum of the masses of
its ingredients," then the dough in the cake weighs 1/6 kilogram.
In 12 pies, ​12*1/6 = 2 kilograms of dough
Solution (B)
Let’s express the amount of collected strawberries as S, and amount of blueberries as B.
We know Little Red Riding Hood’s productivity in collecting blueberries. Her productivity in
collecting strawberries is 1.5 more than that.
So instead of 1 unit of blueberry, she can collect 1.5 units of strawberries.
Little Red Riding Hood can collect 8⋅1.5=12 kilograms of strawberries (at maximum).
Collecting B units of blueberries, Little Red Riding Hood gives up the opportunity to collect
1.5*B units of strawberries, so she collects S=12–1,5B.
Proportions of ingredients in a pie: S/B=(1/2)/(1/3) ⇔ S/B=3/2 ⇔ S=1,5B.
Let’s equate the functions:
1,5B=12–1,5B ⇔ 3B=12 ⇔
B=4 ⇒
S=6
(S/(½)=6/(½)=12)
So there will be 12 pies.
The girl will collect 6 kg of strawberries, and 4 kg of blueberries.
Since it is said that "the mass of the finished cake is strictly equal to the sum of the masses of
its ingredients," we can find mass of dough in each pit.
Pie consists of 3 parts and weighs 1 kg.
Each pie weighs 1 kg and ½ of it is strawberries, ⅓ of it is blueberries, and some part is dough.
Strawberries and blueberries weigh 1/2+1/3=5/6 kg.
Therefore, in each pie we have ⅙ kg of dough, and in total amount of pies – 1/6⋅12=2 kg of
dough.
Problem 3
1. PPF is given by the equation: y = 18−3x;
Find an opportunity cost of :
1. The third unit of x
2. Three units of x
3. Third unit of y
4. Three units of y
2. PPF is given by the equation: y = 100−2 x2 ;
Find an opportunity cost of:
1.
2.
3.
4.
Fifth unit of x
Five units of x
Fifth unit of y
Five units of y
3. PPF is given by the equation: y =
Find an opportunity cost of:
1. Sixth unit of x
√100 − x 2 ;
2. Six units of x
3. Sixth unit of y
4. Six units of y
Solutions
1. ​y=18−3x
1.1
Let’s find the Opportunity Cost OCx at point x=3
y′(x)=−3
y′(3)=−3
OC of good x is constant and equal to 3 units of good y
1.2
Now we are at the point (3;9)
Opportunity Cost=FO−CO
where:
FO=Return on best foregone option
CO=Return on chosen option
Maximum amount of production of good y is equal to 18
Now we find the difference between Maximum amount of production of good y (18) and
Actual amount of produced good at this point (9) x=3.
18−9=9.
OC of 3 units of good x is equal to 9 units of good y
1.3
Let’s do the same for good y
We express x through y
y=18−3x
3x=18−y
x=6−y/3
Find the derivative of this function
x′(y)=−1/3
OC of good y is constant and equal to 1/3 units of good x
1.4
If we produce 3 units of good y, we may produce only 5 units of good x at maximum
In general, we can produce 6 units of good x, when y=0.
6−5=1
OC of 3 units of good y is equal to 1 unit of good x
2.​ y=100−2 x2
2.1
y′(x)=−4x
OC of the 5th unit of x = 4⋅5 = 20
OC of the 5th unit of x is equal to 20 units of good y.
2.2
ymax=100 (when x=0)
y(5)=100−2⋅5^2=50
100−50=50
OC of 5 units of good x is equal to 50 units of good y
OC of the 5th unit of y is equal to
units of good x.
2.4
xmax=5√2 (approx. 7.07)
x(5)=√47,5 (approx 6.89)
OC of 5 units of good y is equal to 7,07−6,89=0,18 units of good y
OC of the 6th unit of good x is equal to 0.75 units of good y
3.2
We can produce 10 units of good y in total.
If
x=6, y=8.
10−8=2
OC of 6 units of good x is equal to 2 units of good y
3.3
Let’s express y using x:
This is the same as the original function, so answers are identical
OC of the 6th unit of good y is equal to 0.75 units of good x
OC of 6 units of good y is equal to 2 units of good x