MATH 1131 ASSIGNMENT
Economics Flavour
z5478989
Intermediate Value Theorem
Q2) Show that the equation
𝑒 −9𝑥 + 7cos(31x) = 0
𝜋
has a unique solution for x ∈ [0, 31].
SOLUTION:
The Intermediate Value Theorem states that for a continuous function 𝑓(𝑥) bounded in a closed
interval[𝑎, 𝑏], if 𝑘 is a given value that lies between 𝑓(𝑎) and 𝑓(𝑏), then there exists at least one real
number c in [𝑎, 𝑏], such that 𝑓(𝑐) = 𝑘.
We must use the Intermediate Value Theorem to show there exists a solution for the equation. Note
that it does not guarantee a unique solution.
𝑓(𝑥) = 𝑒 −9𝑥 + 7 cos(31𝑥)
Since IVT only applies on continuous and differentiable functions for all real 𝑥 , we know that the
individual components of 𝑓(𝑥), 𝑒 −9𝑥 and 7 cos(31𝑥) are both differentiable and continuous.
Hence, we can conclude that 𝑒 −9𝑥 + 7 cos(31𝑥) is continuous on any closed interval, including the
𝜋
given interval x ∈ [0, 31]
For the function at its left most point 𝑥 = 0 ,
𝑓(0) = 𝑒 −9(0) + 7 cos(31 × 0)
= 𝑒 0 + 7 cos(0)
=1+7
=8
>0
For the function at its right-most point 𝑥 =
𝑓(
𝜋
,
31
𝜋
𝜋
𝜋
) = 𝑒 −9(31) + 7 cos (31 × )
31
31
9𝜋
= 𝑒 − 31 + 7 cos(𝜋)
= 0.4017 − 7
≈ −6.5983 …
<0
𝜋
𝜋
Since 𝑓(0) > 0 and 𝑓 (31) < 0, the IVT states that a real number must exist for 𝑐 ∈ [0, 31], where
𝑐 ∈ ℝ such that 𝑓(𝑐) = 0.
𝜋
Now we must prove the function to be strictly monotonic in the interval 𝑐 ∈ [0, 31] to show that
there is a unique solution. We can achieve this by deriving the equation.
𝑓(𝑥) = 𝑒 −9𝑥 + 7 cos(31𝑥)
𝑓 ′ (𝑥) = − 9𝑒 −9𝑥 − 217 sin(31𝑥)
𝜋
We know that for 𝑥 ∈ [0, 31],
𝑒 −4𝑥 > 0
9𝑒 −4𝑥 > 0
−9𝑒 −4𝑥 < 0
Similarly,
7 cos(31𝑥) > 0
217 cos(31𝑥) > 0
−217 cos(31𝑥) < 0
By adding these two components together, we get,
−9𝑒 −4𝑥 − 217 cos(31𝑥) < 0
𝜋
Therefore, 𝑓 ′ (𝑥) < 0 for the interval [0, 31]. This implies that 𝑓(𝑥) is strictly monotonically
decreasing within this interval.
Since the function is monotonically decreasing, there exists a unique solution for 𝑓(𝑥) = 𝑒 −9𝑥 +
7 cos(31𝑥) where 𝑥 ∈ [0,
𝜋
31
END OF Q2 SOLUTIONS.
]as shown by the Intermediate Value Theorem.
The Mean Value Theorem
2
2
Q3) If you don’t have a calculator, you may want to approximate (8.022) 3 by 83
Use the Mean Value Theorem to estimate the error in making this approximation.
SOLUTION:
The Mean Value Theorem states that if 𝑓(𝑥) is continuous on [𝑎, 𝑏], and differentiable on (𝑎, 𝑏), then
there is at least one real number c in (𝑎, 𝑏) such that,
𝑓 (𝑏 ) − 𝑓(𝑎 )
= 𝑓 ′ (𝑐)
𝑏−𝑎
Consider the function,
2
Let 𝑓(𝑥) = 𝑥 3 , 𝑐 ∈ [8,8.022]
which is both continuous and differentiable for [8,8.022].
As a result, we can now substitute values of 𝑎 = 8 and 𝑏 = 8.022 into the Mean Value Theorem
equation as shown,
2
2
(8.022)3−(8)3
= 𝑓 ′ (𝑐)
8.022−8
2
(8.022) 3 − 4
= 𝑓 ′ (𝑐)
0.022
To solve 𝑓 ′ (𝑐), we first need to differentiate 𝑓(𝑥).
2
𝑓(𝑥) = 𝑥 3
𝑓 ′ (𝑥) =
2 −1
𝑥 3
3
2 1
∴ 𝑓 ′ (𝑐) = 𝑐 −3
3
To solve for most accurate error margin, we need to solve for largest value of 𝑓 ′ (𝑐) by substituting
𝑐 = 8 for 𝑐 ∈ [8,8.022]. This is when the error is maximised.
𝑓 ′ (8) =
=
1
2
(8)−3
3
1
3
∴ 𝑓 ′ (𝑐) <
1
3
Combining the two equations of 𝑓 ′ (𝑐) we have,
2
(8.022) 3 − 4
= 𝑓 ′ (𝑐)
0.022
2
(8.022) 3 − 4 1
<
0.022
3
Rearranging the equation,
2
(8.022)3 − 4 <
2
1
× 0.022
3
(8.022)3 − 4 <
11
1500
Hence, the Mean Value Theorem shows that the magnitude of the approximate error by
2
11
approximating 83 is less than 1500.
END OF Q3 SOLUTIONS.
The Intermediate Value/Minimum and Maximum Theorem
Q4) A farm produces cheese and sells it by weight.
The corresponding revenue function is
𝑅(𝑥) = 96𝑥
and the daily cost function is
𝐶(𝑥) = 40 + 146𝑥 0.86
Where 𝑅(𝑥) and 𝐶(𝑥) are in euros and 𝑥 denotes the weight of cheese sold, in kilograms.
For practical reasons (milk supply, storage, and display space), the maximum amount of cheese
that the farm can produce and sell daily is 31.0 kilograms.
a) State an interval of length at most 10 grams that contains a break-even point for the sales of
cheese, where the break-even point is expressed in kilograms. Explain how you know this
interval contains a break-even point.
SOLUTIONS:
The Intermediate Value Theorem states that for a continuous function𝑓(𝑥) bounded in a closed
interval [𝑎, 𝑏], if 𝑘 is a given value that lies between 𝑓(𝑎) and 𝑓(𝑏), then there exists at least one real
number c in [𝑎, 𝑏] , such that 𝑓(𝑐) = 𝑘.
Given,
Revenue function: 𝑅(𝑥) = 96𝑥
Cost function: 𝐶(𝑥) = 40 + 146𝑥 0.86
where x = weight of cheese in kilograms.
We can solve the break-even point by subtracting 𝐶(𝑥) from 𝑅(𝑥). Let the break-even point be
𝑃(𝑥) = 0
𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥) = 0
𝑃(𝑥) = 96𝑥 − 40 − 146𝑥 0.86
∴ 96𝑥 − 40 − 146𝑥 0.86 = 0
By solving for x, we can identify the interval of the break-even point.
𝑥 = 22.79365567
Hence, the interval 𝐼 can be written as,
𝐼: [22.79,22.80] → ℝ for 𝑃(𝑥) = 96𝑥 − 40 − 146𝑥 0.86
We substitute the values of closed bounds, as such,
𝑃(22.79) = 96(22.79) − 40 − 146(22.79)0.86
≈ −0.0546 …
<0
And,
𝑃(22.80) = 96(22.80) − 40 − 146(22.80)0.86
≈ 0.09485 …
>0
Since 𝑓(22.80) > 0 and 𝑓(22.79) < 0, the IVT states that a real number must exist for 𝑐 ∈
[22.79,22.80], where 𝑐 ∈ ℝ such that 𝑃(𝑐) = 0
Therefore, there exists a break-even point that lies between the intervals [22.79,22.80] such that
𝑃(𝑐) = 0.
END OF QUESTION 4(a) SOLUTIONS
b) Prove that the sale of cheese will produce a maximum profit. Hence, find how many
kilograms of cheese should be sold to maximise profit.
In addition, to avoid the worst possible financial situation, the farmers would also like to know
selling how many kilograms of cheese will result in the greatest loss. In order to answer this
question, find how many kilograms of cheese sold will minimise profit.
SOLUTIONS:
The Maximum-Minimum Theorem states that if 𝑓 is continuous on a closed interval [𝑎, 𝑏] then
𝑓 attains its minimum and maximum on [𝑎, 𝑏].
Profit is calculated by the difference between Revenue and Cost,
𝑃(𝑥) = 𝑅(𝑥) − 𝐶(𝑥)
𝑃(𝑥) = 96𝑥 − 40 − 146𝑥 0.86
Given the minimum and maximum amount of cheese that can be sold per day, we can deduce that,
𝑃: [0,31] → 𝑅 for 𝑃(𝑥) = 96𝑥 − 40 − 146𝑥 0.86
To take both minimum and maximum value on [0,31] of 𝑃(𝑥), we must derive 𝑃(𝑥) to assess the
points of inflexion and its concavity to determine maxima or minima points of 𝑃(𝑥).
𝑃(𝑥) = 96𝑥 − 40 − 146𝑥 0.86
𝑃′ (𝑥) = 96 − 125.56𝑥 −0.14
Let 𝑃′ (𝑥) = 0 to solve for 𝑥
0 = 96 − 125.56𝑥 −0.14
96 = 125.56𝑥 −0.14
𝑥 −0.14 =
2400
3139
1
𝑥=
100
√(2400)
3139
14
= 6.803224683
≈ 6.803 (To the nearest gram)
To find the concavity of 𝑃(𝑥), we must find the second derivative as such,
𝑃 (𝑥) = 96𝑥 − 40 − 146𝑥 0.86
𝑃′ (𝑥) = 96 − 125.56𝑥 −0.14
𝑃′′ (𝑥) = 17.5784𝑥 −1.14
By substituting 𝑥 = 6.803 ,
𝑃′′ (6.803) = 17.5784(6.803)−1.14
= 1.975608135
>0
Since 𝑃′′ (𝑥) > 0 at 𝑥 = 6.803 for 𝑥 ∈ [0,29], 𝑃(𝑥) concaves upwards and has a local minimum
turning point.
We know that 𝑥 = 6.803 is the minimum point in 𝑃(𝑥) as 𝑃(𝑥) decreases when 𝑥 < 6.803 and
increases when 𝑥 > 6.803 . Hence, we can conclude that 6.803 Kg of the sale of cheese is the lowest
profit.
Similarly, taking the closed points of [0,31]
𝑃(0) = 96(0) − 40 − 146(0)0.86
= −40
<0
𝑃(31) = 96(31) − 40 − 146(31)0.86
= 137.5090133
>0
Hence, local maximum is at 𝑥 = 31 since 𝑃(31) > 𝑃(0). We can conclude that the sale of cheese of
31kg is when profit is maximised.
Therefore, profit is maximised at 31.000kg of cheese sold, whereas the greatest loss of profit occurs
when 6.803kg of cheese are sold.
END OF QUESTION 4(b) SOLUTIONS
END OF TASK