QUESTION 1
Calculus can be employed to optimize the dimensions of a cylinder and demonstrate
that for a right circular cylinder with a specified surface area, including the ends, and the
maximum possible volume, its height should be equal to the diameter of its base. This
involves formulating and solving an optimization problem.
Let: A = Surface area of the cylinder,
D = Diameter of the cylinder,
H = Height of the cylinder,
V = Volume of the cylinder,
The total surface area (including the ends) of a right circular cylinder is given by:
𝜋𝐷2
𝐴 = 2(
) + 𝜋𝐷𝐻
4
𝜋𝐷2
𝐴=
+ 𝜋𝐷𝐻
2
The volume of the cylinder is given by:
𝜋𝐷2
𝑉=(
)𝐻
4
Now, we are given a fixed surface area A, and we want to maximize the volume V. We
can set up problem as follows:
𝜋𝐷 2
Maximize, 𝑉 = (
4
)𝐻
Subject to the constraint, 𝐴 =
𝜋𝐷 2
2
+ 𝜋𝐷𝐻
We can now solve for H in terms of D from the constraint equation:
𝐴=
𝜋𝐷2
+ 𝜋𝐷𝐻
2
𝜋𝐷2
𝜋𝐷𝐻 = 𝐴 −
2
𝐻=
𝐴
𝐷
−
𝜋𝐷 2
Now we can substitute this expression for H into the volume equation:
𝜋𝐷2
𝐴
𝐷
𝑉=(
)(
− )
4
𝜋𝐷 2
Now simplify the expression of V:
𝑉=
𝐴
𝜋
𝐷 − 𝐷3
4
8
To maximize V, we take the derivative of V with respect to D and set it equal to zero:
𝑑𝑉
=0
𝑑𝐷
It is important to note that the surface area is constant, and therefore, the derivative of V
is:
𝑑𝑉 𝐴 3𝜋 2
= −
𝐷
𝑑𝐷 4
8
Now set the derivative equal to zero:
𝐴 3𝜋 2
−
𝐷 =0
4
8
Now solve for A:
𝐴 3𝜋 2
−
𝐷 =0
4
8
𝐴=
3 2
𝜋𝐷
2
Now let substitute this A expression on the original surface area equation:
𝐴=
Original equation:
𝜋𝐷 2
2
+ 𝜋𝐷𝐻
3𝜋𝐷2 𝜋𝐷2
=
+ 𝜋𝐷𝐻
2
2
Now let’s solve the expression and see if the diameter is equal to the height of the
cylinder at maximum volume V:
Dividing with 𝜋𝐷:
3𝐷 𝐷
= +𝐻
2
2
Grouping the like-terms:
(
3𝐷 𝐷
− )=𝐻
2
2
𝐷=𝐻
Therefore, the right circular cylinder with the given surface (including the ends) and
maximum volume has a height equal to the diameter of the base.
QUESTION 2
i)
To maximize profit, the manufacturer needs to schedule the production of chair
and tables on the three machines in a way that maximizes the total profit while
respecting the time available on each machine. This can be solved as a linear
programming problem. We need to formulate the problem mathematically and
then use an optimization technique to find the solution:
Let: X = Chairs
Y = Tables
With the individual chair has the profit of $20 and the table has $30, using linear
programming method, the profit can be calculated as:
𝑃 = 20𝑋 + 30𝑌
Where: P = Profit of the firm, $
The provided data indicates that Machine 1 (M1) can manufacture one chair and one
table in 3 hour, and it is available for 36 hours per week. Machine 2 (M2) can produce a
maximum of 5 chairs and 2 tables within 50 hours a week, while Machine 3 (M3) can
produce up to 2 chairs and 6 tables in 60 hours during the week. These specifications
serve as constraints for the firm.
Constrains:
M1: 3𝑋 + 3𝑌 ≤ 36
M2: 5𝑋 + 2𝑌 ≤ 50
M3: 2𝑋 + 6𝑌 ≤ 60
The produced chairs and table can never be negative; they can only be zero or positive
number. Therefore, there are new constrains, which is:
𝑋≥0
𝑌≥0
To optimize profit, the scheduling of production was assessed using Excel's Solver,
yielding the subsequent findings.
Microsoft Excel 14.0 Answer Report
Worksheet: [New Microsoft Excel Worksheet.xlsx]Sheet1
Report Created: 10/20/2023 10:39:23 PM
Result: Solver found a solution. All Constraints and optimality conditions are
satisfied.
Solver Engine
Engine: GRG Nonlinear
Solution Time: 0.406 Seconds.
Iterations: 3 Subproblems: 0
Solver Options
Max Time Unlimited, Iterations Unlimited, Precision 0.000001, Use Automatic
Scaling
Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Forward,
Require Bounds
Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%,
Assume NonNegative
Objective Cell (Max)
Cell
Name
Original Value
$B$6
P
50
Variable Cells
Cell
Name
$B$2
X
$B$3
Y
Constraints
Cell
Name
$C$12
$C$13
$C$14
$C$15
$C$16
Original Value
Final Value
330
Final Value
Integer
3 Contin
9 Contin
Formula
$C$12<=$D$12
$C$13<=$D$13
$C$14<=$D$14
$C$15>=$D$15
$C$16>=$D$16
Status
Binding
Not Binding
Binding
Not Binding
Not Binding
1
1
Cell Value
36
33
60
1
1
Slack
0
17
0
1
1
ii)
DISCUSSION
By utilizing Solver, an optimal production schedule was determined for the
manufacturer. It was recommended that Machine 1 (M1) should operate for 36 hours
per week, Machine 2 (M2) should run for 33 hours per week, and Machine 3 (M3)
should be in operation for 60 hours a week.
Upon analyzing the provided data, it became evident that tables generate higher profits
than chairs. Consequently, to maximize profitability, it is advisable to prioritize the
production and sale of tables. Further examination of the machine data revealed that
Machine 2 produces more chairs than tables, Machine 1 produces an equal number of
chairs and tables each week, and Machine 3 produces three times as many tables as
chairs weekly. As a result, it is reasonable to minimize the operation of Machine 2 and
give preference to Machines 1 and 3.
The Solver optimization solution indicated that the maximum profit could be achieved by
maintaining a sales ratio of 3 chairs to 9 tables, resulting in a profit of $330.