Practical problems
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Practical problems
ELEMENTARY MATHEMATICS AND APPLICATIONS OF CALCULUS
INGRIDA VEILANDE LATVIAN MARITIME ACADEMY
Innovative Approach in Mathematical Education for Maritime Students
2019-1-HR01-KA203-061000
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Problem 1.
The farmer has a lot of hens and chickens. They are
growing naturally in the open air. The farmer decides to
construct the chicken yard. He has 300 m of wire gaze and he
wants to construct the rectangular yard with the maximal
possible area. Calculate the size of this yard!
x
y
x
x
y
Perimeter
P = 2x + 2y = 300 m
Area
Function
Solution of problem 1
Algebraic calculation
y = 150 – x = 75
Analytic method
S’(x) =150 - 2x
Problem 2.
The rectangular yard
must be constructed at the
wall using 300 m of wire gaze for three sides of this yard. Find
the lengths of the edges that the yard has a maximal area.
Solution of problem 2
x
x
y
Perimeter
of the
wire gaze
P = 2x + y = 300
Area
of the yard
Function
Calculation
S’(x) =300 - 4x
Problem 3.
From the metallic sheet of size 2 x 2 m must be
fabricated the container with maximal volume. The master will
cut out the squares at the corners of metallic sheet and he will
solder the edges. What size will have the squares?
x
Side of the corner’s square x
Side of the bottom square 2 - 2x
Volume of the container
Solution of problem 3
Graph of the
function V(x)
x
Maximal value of the function is at the point x = c
Value V(x) < V(c) for all
We will investigate similar function F(x)
Maximal value of the expression is when x = c.
It gives
Answer:
Problem 4.
For practical reasons the fitter has to fabric the steel
plate of size 10 x 10 cm with a rectangular hole in the centre.
Around the hole must be solder a wire. The weight of 1 cm
2
of the
plate is 2 g, the weight of 1 cm of the wire is 7 g. The diagonal of
the rectangular hole is 5 cm long. The challenge is to make this
detail of the maximal possible weight.
Sides of hole x and y
Diagonal d = 5 cm
d
d
The weight of the plate
The weight of the catted out part
2xy
The weight of wire
The weight of detail
The lengths of diagonal
Solution of problem 4
To express the weight as a function with respect of one
argument we will do algebraic transformations
By the substitution
t = x + y
Let us derivate the function
At the
t=7
the function
has their maximal value
To find the size of the hole
The solution of the system
d
Answer:
(x; y) = (3; 4) or (x; y) = ( 4; 3)
The detail has their maximal weight if the hole
is of size 3 x 4 cm;
W = 274
g
B
A
O
l
B
A
O
l
The angle of the sector in radians
The lengths of the arc
The radius of the correspondent circumference
r = OB
Solution of problem 5
/Because of
Let us investigate the function
The derivative
To find the critical points of the derivative
The graph of the
function
The critical points for the derivative
are when
We get
Answer:
the area of the segment is
Problem 6.
A fence
q
metre tall runs parallel to a building at a
distance of
p
metre from the building. What is the length of
the shortest ladder AD that can
reach from the ground over the
fence to the wall of the building?
By the theorem of Pythagoras
where
OA = p + a
and
OD = q + ED
From the similarity of triangles
ABC
and
BDE
or
Solution of problem 6
The length of ladder
We will find the minimal value of the function
Let us derivate the function with respect to
a
Extreme value is when
If
If
then
then
The minimal value of
a
is
Answer:
The length of the shortest
ladder at the variable
a
Problem 7.
Find the area of the region
The region is a part of the circle
The area is
Calculation by substitution
Answer:
The area of the region is
Geometric approach
In the triangle ABC
The area of the region is a
difference
Explore a new approach to mathematical education for maritime students with practical problems in elementary mathematics and applications of calculus. Dive into solving problems involving maximizing areas of chicken yards, finding optimal yard dimensions, and fabricating containers with maximum volume. Enhance your mathematical skills with real-world scenarios designed for maritime studies.
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Presentation Transcript
Innovative Approach in Mathematical Education for Maritime Students 2019-1-HR01-KA203-061000 Practical problems ELEMENTARY MATHEMATICS AND APPLICATIONS OF CALCULUS INGRIDA VEILANDE LATVIAN MARITIME ACADEMY
Problem 1. The farmer has a lot of hens and chickens. They are growing naturally in the open air. The farmer decides to construct the chicken yard. He has 300 m of wire gaze and he wants to construct the rectangular yard with the maximal possible area. Calculate the size of this yard! x y
Solution of problem 1 Perimeter P = 2x + 2y = 300 m x Area ? = ? ? = ? (150 ? Function y ?(? = 150? ?2
Algebraic calculation ? ? = 150? ?2= ?2 150? = = (?2 2 75? + 752 752 = = ((? 75 2 752 = 752 (? 75 2 y = 150 x = 75 Answer:????= ? ? = 752= 5625?2
Analytic method ?(? = 150? ?2 S (x) = 0 S (x =150 - 2x 150 2? = 0; ? = 75 ????= ? ? = 752= 5625?2
Problem 2. The rectangular yardmust be constructed at the wall using 300 m of wire gaze for three sides of this yard. Find the lengths of the edges that the yard has a maximal area.
Solution of problem 2 Perimeter of the wire gaze x x P = 2x + y = 300 y Area of the yard ? = ? ? = ? (300 2? ?(? = 300? 2?2 Function
Calculation ?(? = 300? 2?2 S (x)= 0 S (x =300 - 4x ? = 75 ??????: ? 75 = 300 75 2 752= 75 150 = 11250 ?2 Answer to the previous problem 1 S = 5625 ?2
Problem 3. From the metallic sheet of size 2 x 2 m must be fabricated the container with maximal volume. The master will cut out the squares at the corners of metallic sheet and he will solder the edges. What size will have the squares?
Solution of problem 3 x Side of the corner s square x Side of the bottom square 2 - 2x Volume of the container 2 2?2? ?(? =
Graph of the function V(x) 2 2?2? = ? ? = = 4?3 8?2+ 4? x
2 2?2? ?(? = Maximal value of the function is at the point x = c Value V(x) < V(c) for all ? [0; 1] ??? ? ? We will investigate similar function F(x) ?(? = (1 ? 2? = ? 2?2 + ?3 ?(? ?(?
If ?(? ?(? then ?3 2?2 + ? ?3 2?2 + ? ?3 ?3 2(?2 ?2 + (? ? 0 (? ? (?2 + ?? + ?2 2? 2? + 1 0 Maximal value of the expression is when x = c. It gives 3?2 4? + 1 = 0 ? =16 ? =1 3; Answer: 27 0.5926
Problem 4. For practical reasons the fitter has to fabric the steel plate of size 10 x 10 cm with a rectangular hole in the centre. Around the hole must be solder a wire. The weight of 1 cm 2 of the plate is 2 g, the weight of 1 cm of the wire is 7 g. The diagonal of the rectangular hole is 5 cm long. The challenge is to make this detail of the maximal possible weight. Sides of hole x and y Diagonal d = 5 cm d
Solution of problem 4 2 100 ? The weight of the plate d The weight of the catted out part 2xy 7 (2? + 2? The weight of wire ? = 200 2?? + 14(? + ? The weight of detail ?2= ?2+ ?2=25 The lengths of diagonal
To express the weight as a function with respect of one argument we will do algebraic transformations ? = 200 2?? + 14(? + ? ?2+ ?2= ?2+ ?2+ 2?? 2?? = ? + ?2 2?? = 25 ?(?,? = 200 2?? + 14 ? + ? = 200 ( ? + ?2 25 + 14(? + ? By the substitution t = x + y ?(? = 200 (?2 25 + 14? = ?2+ 14? + 225
?(? = ?2+ 14? + 225 Let us derivate the function ? (? = 2? + 14 At the t=7 the function has their maximal value ?(? = 49 + 14 7 + 225 = 274 ? ? + ? = 7 ?2+ ?2= 25 To find the size of the hole
? + ? = 7 ?2+ ?2= 25 The solution of the system ? + ? = 7 2?? = (? + ?2 25 ? + ? = 7 ?? = 12 d Answer: (x; y) = (3; 4) or (x; y) = ( 4; 3) The detail has their maximal weight if the hole is of size 3 x 4 cm; W = 274 g
Problem 5. We want to conduct water through a bended rectangular piece of metal so that the section of the piece is an arc of circumference. Calculate the value of the angle that the shadowed region ?(? is maximal (and therefore the amount of water through it). O A B S( ) l
Solution of problem 5 ??? = The angle of the sector in radians 0 < 2? The radius of the correspondent circumference r = OB The lengths of the arc ? = ? O The area of the segment ??? ? ? = ? ????????? ?( ??? A B S( ) ? ? =? 2?2 1 2?2???? l
? ? =? 2?2 1 2?2???? = ?2 2?2? ???? = = ?2 ? ???? = ? = ? / /Because of 2 =?2 2 ? ???? ?2 Let us investigate the function ? ? =? ???? ?2
? ? =? ???? ?2 1 ???? ?2 2?(? ???? ?4 = ?( 1 ???? ? 2(? ???? ?4 1 ???? ? 2(? ???? ?3 = 2???? ? 1+???? ?3 The derivative ? ? = = = = =
To find the critical points of the derivative S (x)= 2???? ? 1+???? = ?3 2 2???? 2???? 2 ? 2???2? ?3 2 = = 4???2? (???? 2 ? 2 2 = ?3
The graph of the function ? 2 ? ? ? = tan 2 ? ? > 0 ?? 0 < ? < ? ? ? ?? ??? ??????? ?? ? = ?
The critical points for the derivative 4???2? (???? 2 ? 2 2 S (x) = ?3 ???2? are when 2 = 0 for 0 < x 2? We get x = ? Answer: the area of the segment is 2?2? ???? =?2 ?2 ? ? = 2?
Problem 6. A fence q metre tall runs parallel to a building at a distance of p metre from the building. What is the length of the shortest ladder AD that can reach from the ground over the fence to the wall of the building?
Solution of problem 6 By the theorem of Pythagoras ??2= ??2+ ??2 where OA = p + a and OD = q + ED From the similarity of triangles ABC and BDE ?? ? =? ?? ??=?? or ? ??
The length of ladder (? +?? ? 2+(? + ? 2 ? ? = ?? = We will find the minimal value of the function f(a)=(? +?? ? 2+(? + ? 2
Let us derivate the function with respect to a f(a)=(? +?? ? 2+(? + ? 2 ? (? =2(q+?? ? ?? ?2+ 2 ? + ? = =2q(a + p) ?? ?2+2 ? + ? = =2(p + a)(1 -??2 ?3
Extreme value is when (1 -??2 ?3 = 0 ??2 ?3=1 3??2 a = (1 -??2 3??2 then a < ?3 < 0 If (1 -??2 3??2 a > If then ?3 > 0
3??2 a = The minimal value of a is Answer: The length of the shortest ladder at the variable a ? ? = ? ? = 3?2? 2+(? + 3??2 2 = (? +
Problem 7. Find the area of the region (?,? ? 0,? 1,?2+ ?2 4? ? = The region is a part of the circle ?2+ (? 2 2= 4 The area is 1 4? ?2?? ? = 0
Calculation by substitution 1 4? ?2?? ? = 0 1 1 4? ?2?? = 4 (? 2 2?? 0 ? 2 = 2????;?? = 2??????; ?1= ? 0 2; ?2= ? ?6(1 + ???2? ?? 6 ?6 4 4???2? 2?????? = 2 ?2 ?2
?6 ?2 ?6 2???2? 1 + ???2? ?? = 2(? +1 = 2 = ?2 ? 3 1 3 =2? 3 = 2 2 3 2 2 Answer: The area of the region is S =2? 3 3 2
Geometric approach In the triangle ABC ??? = 60 =? 3 The area of the region is a difference ? = ????????? ??? ? =1 2 22 ? 3 1 Answer: S =2? 3 3 2 1 3 2
