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Question 1
The surfing club provides swimming lessons to boost their funds. The club has
fixed costs of R1035 per day, when offering these classes-mostly for
insuranceand it costs them R195 for each lesson given. How many lessons did
they provide if total costs amounted to R3000 ?
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Solution 1
The surfing club’s total costs function is
TC = FC + V C = 1035 + 195q,
where q is the number of lessons given per day.
When the club’s total costs is R3000 then the number of lesson will be
Question 2
A firm’s fixed cost to produce cups amount to R1260 per week and it costs R22
to produce each cup. These cups are sold for R65 each. What would be the firm’s
profit if they produce and sell 160 cups?
Solution 2
profit(P) = TR − TC
where
TR = price per cup × number of cups sold
= 65q,
where q is the number of cups produced (or sold) per week.
TC = FC + V C
= 1260 + 22q.
Hence, the profit function becomes
P = 65q − (1260 + 22q)
= 65q − 1260 − 22q
= 43q − 1260.
Also, when they sell 160 cups, their profit is R5620, which can be calculated as
P(160) = 43(160) − 1260 = 5620.
Question 3
The demand and supply functions for pans are
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q = 500 − 5p
and p = 0,8q + 30,
respectively, with the price in rand and q the quantity in boxes. Find the point(s)
where the demand and supply function intersect.
Solution 3
The demand function q = 500 − 5p can be transformed to
Now, that both the demand and supply function are in the form p(q), we can
equate them to find the point where they intesect.
100 = 0,2q = 0,8q + 30
100 − 30 = 0,2q
+ 0,8q q = 70.
To find the value of p, we substitute q = 70 on one of the given equations. Then
p = 100 − 0,2(70) = 86.
Question 4
The demand function for a certain kind of cellphones is given by
q = 90000 − 5p.
Determine the point elasticity of demand when p = 5000.
Solution 4
When p = 5000, then
q = 90000 − 5(5000) = 65000,
that is (p0;q0) = (5000;65000) and
|ϵ| = 0,385 < 1, the demand is inelastic at this price. 1% increase (or decrease) in
price will cause a 0,385% decrease (or increase) in demand.
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Question 5
Consider the supply function for a kind of computer, namely
p = 10 + 0,5q.
Determine the arc elasticity if price increases from R1000 to R3000.
Solution 5
The supply function p = 10 + 0,5q can be transformed to q = 2p − 20.
We need to find the quantities demanded at these prices.
If p = 1000, then q = 2(1000) − 20 = 1980, to get (p1;q1) = (1000;1980).
If p = 3000, then q = 2(3000) − 20 = 5980, to get (p1;q1) = (3000;5980).
The arc price elasticuty is
|ϵ| = 1, the supply is unit elastic. The 1% change in supply is equal to 1% change
in price.
Question 6
The demand and supply functions for a certain product are given as
qd = 300 − 2pd
and
qs = −200 + 3ps,
respectively. Determine the price (p) and quantity (q) at equilibrium.
Solution 6
To find the equilibrium price and quantity, we need to solve the demand and
supply functions simultaneously. Since the functions are both given in the form
q(p), we set them equal and solve for p (the equilibrium price), that is
qd = qs
300 − 2p = −200 + 3p
300 + 200 = 2p + 3p
500 = 5p
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Question 1
The surfing club provides swimming lessons to boost their funds. The club has
fixed costs of R1035 per day, when offering these classes-mostly for
insuranceand it costs them R195 for each lesson given. How many lessons did
they provide if total costs amounted to R3000 ?
1
Downloaded by Vincent kyalo ()
, lOMoARcPSD|48011787
Solution 1
The surfing club’s total costs function is
TC = FC + V C = 1035 + 195q,
where q is the number of lessons given per day.
When the club’s total costs is R3000 then the number of lesson will be
Question 2
A firm’s fixed cost to produce cups amount to R1260 per week and it costs R22
to produce each cup. These cups are sold for R65 each. What would be the firm’s
profit if they produce and sell 160 cups?
Solution 2
profit(P) = TR − TC
where
TR = price per cup × number of cups sold
= 65q,
where q is the number of cups produced (or sold) per week.
TC = FC + V C
= 1260 + 22q.
Hence, the profit function becomes
P = 65q − (1260 + 22q)
= 65q − 1260 − 22q
= 43q − 1260.
Also, when they sell 160 cups, their profit is R5620, which can be calculated as
P(160) = 43(160) − 1260 = 5620.
Question 3
The demand and supply functions for pans are
2
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q = 500 − 5p
and p = 0,8q + 30,
respectively, with the price in rand and q the quantity in boxes. Find the point(s)
where the demand and supply function intersect.
Solution 3
The demand function q = 500 − 5p can be transformed to
Now, that both the demand and supply function are in the form p(q), we can
equate them to find the point where they intesect.
100 = 0,2q = 0,8q + 30
100 − 30 = 0,2q
+ 0,8q q = 70.
To find the value of p, we substitute q = 70 on one of the given equations. Then
p = 100 − 0,2(70) = 86.
Question 4
The demand function for a certain kind of cellphones is given by
q = 90000 − 5p.
Determine the point elasticity of demand when p = 5000.
Solution 4
When p = 5000, then
q = 90000 − 5(5000) = 65000,
that is (p0;q0) = (5000;65000) and
|ϵ| = 0,385 < 1, the demand is inelastic at this price. 1% increase (or decrease) in
price will cause a 0,385% decrease (or increase) in demand.
3
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Question 5
Consider the supply function for a kind of computer, namely
p = 10 + 0,5q.
Determine the arc elasticity if price increases from R1000 to R3000.
Solution 5
The supply function p = 10 + 0,5q can be transformed to q = 2p − 20.
We need to find the quantities demanded at these prices.
If p = 1000, then q = 2(1000) − 20 = 1980, to get (p1;q1) = (1000;1980).
If p = 3000, then q = 2(3000) − 20 = 5980, to get (p1;q1) = (3000;5980).
The arc price elasticuty is
|ϵ| = 1, the supply is unit elastic. The 1% change in supply is equal to 1% change
in price.
Question 6
The demand and supply functions for a certain product are given as
qd = 300 − 2pd
and
qs = −200 + 3ps,
respectively. Determine the price (p) and quantity (q) at equilibrium.
Solution 6
To find the equilibrium price and quantity, we need to solve the demand and
supply functions simultaneously. Since the functions are both given in the form
q(p), we set them equal and solve for p (the equilibrium price), that is
qd = qs
300 − 2p = −200 + 3p
300 + 200 = 2p + 3p
500 = 5p
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