The document discusses how to solve maximization and minimization word problems. It provides an example of maximizing the total area formed by cutting a rope into two pieces to form a square and rectangle. The steps are given as: (1) write the problem as two equations, (2) make one variable the subject of one equation, (3) substitute and take the derivative, (4) set the derivative equal to 0 to find the maximum/minimum. The maximum total area occurs when the side of the square is 1/2 the length of the rope. A second example finds the radius, height, and maximum volume of a cylindrical bucket given its total surface area.
The document discusses solving maximization and minimization word problems using calculus. It provides steps to (1) reduce the problem to equations, (2) rewrite one equation in terms of one variable, and (3) use calculus to find the maximum or minimum value. An example problem is provided where the goal is to maximize the total area of two pieces cut from a rope by forming one into a square and the other into a rectangle. Using the given steps and calculus, the maximum total area is found when the length of the square is 1/2 meters. A second example involves maximizing the volume of a cylindrical bucket given its total surface area.
The document discusses coordinate geometry concepts including the distance formula and midpoint formula. It explains that the distance formula calculates the length of the hypotenuse between two points using Pythagoras' theorem. The midpoint formula averages the x- and y-coordinates of two points to find the midpoint. It also discusses dividing intervals, noting that for a level 2 math exam it is restricted to midpoint divisions in a 1:1 ratio, while an extension 1 exam can involve any ratio for internal or external divisions. Examples are provided to illustrate the concepts.
1. Related rates problems involve differentiating equations with respect to time to determine an unknown rate of change.
2. The general procedure is to define variables, write an equation relating the variables, implicitly differentiate with respect to time, and substitute known values and rates of change to solve for the unknown rate.
3. Examples include finding the rate of change of total area of a spreading pond ripple or the rate of change of radius of a inflating balloon given the rate air is being pumped in.
This document appears to be about AP Calculus and related rates problems. It contains 6 video links that are presumably related to teaching or explaining related rates problems in calculus. The videos likely provide instruction and worked examples for students studying advanced placement calculus related rates questions.
The document discusses maxima and minima, which are the highest and lowest values that a function can reach over a closed interval. It provides examples of finding maxima and minima by taking derivatives and setting them equal to zero. Practical applications are given in chemistry, physics, economics, meteorology, theme park revenue modeling, and space shuttle engineering where maxima and minima help optimize values like temperature, revenue, and pressure resistance.
The document discusses solving maximization and minimization word problems using calculus. It provides steps to take which include: (1) reducing the problem to two equations, one for the quantity to maximize/minimize and one with given information, (2) rewriting the equation to maximize/minimize with only one variable, and (3) using calculus to solve the problem. An example problem is provided and solved step-by-step to find the length of a square that maximizes the total area of two pieces cut from a rope.
The document discusses solving maximization and minimization word problems using calculus. It provides steps to take which include: (1) reducing the problem to two equations, one for the quantity to maximize/minimize and one with given information, (2) rewriting the equation to maximize/minimize with only one variable, and (3) using calculus to solve the problem. An example problem is provided and solved step-by-step to find the length of a square that maximizes the total area of two pieces cut from a rope.
This document discusses how to solve maxima and minima word problems. It explains that problems should be reduced to two equations, one for the quantity being maximized/minimized and one for given information. The equation for maximizing/minimizing should be rewritten with one variable. Calculus is then used to solve the problem by finding the derivative and setting it equal to zero to find critical points. An example problem is included where the maximum area of two shapes made from a rope is found.
The document discusses solving maximization and minimization word problems using calculus. It provides steps to (1) reduce the problem to equations, (2) rewrite one equation in terms of one variable, and (3) use calculus to find the maximum or minimum value. An example problem is provided where the goal is to maximize the total area of two pieces cut from a rope by forming one into a square and the other into a rectangle. Using the given steps and calculus, the maximum total area is found when the length of the square is 1/2 meters. A second example involves maximizing the volume of a cylindrical bucket given its total surface area.
The document discusses coordinate geometry concepts including the distance formula and midpoint formula. It explains that the distance formula calculates the length of the hypotenuse between two points using Pythagoras' theorem. The midpoint formula averages the x- and y-coordinates of two points to find the midpoint. It also discusses dividing intervals, noting that for a level 2 math exam it is restricted to midpoint divisions in a 1:1 ratio, while an extension 1 exam can involve any ratio for internal or external divisions. Examples are provided to illustrate the concepts.
1. Related rates problems involve differentiating equations with respect to time to determine an unknown rate of change.
2. The general procedure is to define variables, write an equation relating the variables, implicitly differentiate with respect to time, and substitute known values and rates of change to solve for the unknown rate.
3. Examples include finding the rate of change of total area of a spreading pond ripple or the rate of change of radius of a inflating balloon given the rate air is being pumped in.
This document appears to be about AP Calculus and related rates problems. It contains 6 video links that are presumably related to teaching or explaining related rates problems in calculus. The videos likely provide instruction and worked examples for students studying advanced placement calculus related rates questions.
The document discusses maxima and minima, which are the highest and lowest values that a function can reach over a closed interval. It provides examples of finding maxima and minima by taking derivatives and setting them equal to zero. Practical applications are given in chemistry, physics, economics, meteorology, theme park revenue modeling, and space shuttle engineering where maxima and minima help optimize values like temperature, revenue, and pressure resistance.
The document discusses solving maximization and minimization word problems using calculus. It provides steps to take which include: (1) reducing the problem to two equations, one for the quantity to maximize/minimize and one with given information, (2) rewriting the equation to maximize/minimize with only one variable, and (3) using calculus to solve the problem. An example problem is provided and solved step-by-step to find the length of a square that maximizes the total area of two pieces cut from a rope.
The document discusses solving maximization and minimization word problems using calculus. It provides steps to take which include: (1) reducing the problem to two equations, one for the quantity to maximize/minimize and one with given information, (2) rewriting the equation to maximize/minimize with only one variable, and (3) using calculus to solve the problem. An example problem is provided and solved step-by-step to find the length of a square that maximizes the total area of two pieces cut from a rope.
This document discusses how to solve maxima and minima word problems. It explains that problems should be reduced to two equations, one for the quantity being maximized/minimized and one for given information. The equation for maximizing/minimizing should be rewritten with one variable. Calculus is then used to solve the problem by finding the derivative and setting it equal to zero to find critical points. An example problem is included where the maximum area of two shapes made from a rope is found.
Slideshare is discontinuing its Slidecast feature as of February 28, 2014. Existing Slidecasts will be converted to static presentations without audio by April 30, 2014. The document informs users that new slidecasts can be found on myPlick.com or the author's blog starting in 2014. However, myPlick proved unreliable, so future slidecasts will instead be hosted on the author's YouTube channel.
The document discusses different methods for factorising expressions:
1) Looking for a common factor and dividing it out of all terms
2) Using the difference of two squares formula (a2 - b2 = (a - b)(a + b))
3) Factorising quadratic trinomials into two binomial factors by identifying the values that multiply to give the constant term and sum to give the coefficient of the linear term.
The document provides information on index laws and the meaning of indices in algebra:
- Index laws state that am × an = am+n, am ÷ an = am-n, and (am)n = amn. Exponents can be added or subtracted when multiplying or dividing terms with the same base.
- Positive exponents indicate a term is raised to a power. Negative exponents indicate a root is being taken. Terms with exponents are evaluated from left to right.
- Examples demonstrate how to simplify expressions using index laws and interpret different types of indices.
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
The document discusses integrating derivatives of functions. It states that the integral of the derivative of a function f(x) is equal to the natural log of f(x) plus a constant. It then provides examples of integrating several derivatives: (i) ∫(1/(7-3x)) dx = -1/3 log(7-3x) + c, (ii) ∫(1/(8x+5)) dx = 1/8 log(8x+5) + c, and (iii) ∫(x5/(x-2)) dx = 1/6 log(x6-2) + c. It also discusses techniques for integrating fractions by polynomial long division and finds
The document discusses logarithms and their properties. Logarithms are defined as the inverse of exponentials. If y = ax, then x = loga y. The natural logarithm is log base e, written as ln. Properties of logarithms include: loga m + loga n = loga mn; loga m - loga n = loga(m/n); loga mn = n loga m; loga 1 = 0; loga a = 1. Examples of evaluating logarithmic expressions are provided.
The document discusses relationships between the coefficients and roots of polynomials. It states that for a polynomial P(x) = axn + bxn-1 + cxn-2 + ..., the sum of the roots equals -b/a, the sum of the roots taken two at a time equals c/a, and so on for higher order terms. It also provides examples of using these relationships to find the sums of roots for a given polynomial.
P
4
3
2
The document discusses properties of polynomials with multiple roots. It first proves that if a polynomial P(x) has a root x = a of multiplicity m, then the derivative of P(x), P'(x), will have a root x = a of multiplicity m-1. It then provides an example of solving a cubic equation given it has a double root. Finally, it examines a quartic polynomial and shows that its root α cannot be 0, 1, or -1, and that 1/
The document discusses factorizing complex expressions. The main points are:
- If a polynomial's coefficients are real, its roots will appear in complex conjugate pairs.
- Any polynomial of degree n can be factorized into a mixture of quadratic and linear factors over real numbers, or into n linear factors over complex numbers.
- Odd degree polynomials must have at least one real root.
- Examples of factorizing polynomials over both real and complex numbers are provided.
The document describes the Trapezoidal Rule for approximating the area under a curve between two points. It shows that the area A is estimated by dividing the region into trapezoids with height equal to the function values at the interval endpoints and bases equal to the intervals. In general, the area is approximated as the sum of the areas of each trapezoid, which is equal to the average of the endpoint function values multiplied by the interval length.
The document discusses methods for calculating the volumes of solids of revolution. It provides formulas for finding volumes when an area is revolved around either the x-axis or y-axis. Examples are given for finding volumes of common solids like cones, spheres, and others. Steps are shown for using the formulas to calculate volumes based on given functions and limits of revolution.
The document discusses different methods for calculating the area under a curve or between curves.
(1) The area below the x-axis is given by the integral of the function between the bounds, which can be positive or negative depending on whether the area is above or below the x-axis.
(2) To calculate the area on the y-axis, the function is solved for x in terms of y, then the bounds are substituted into the integral of this new function with respect to y.
(3) The area between two curves is calculated by taking the integral of the upper curve minus the integral of the lower curve, both between the same bounds on the x-axis.
The document discusses 8 properties of definite integrals:
1) Integrating polynomials results in a fraction.
2) Constants can be factored out of integrals.
3) Integrals of sums are equal to the sum of integrals.
4) Splitting an integral range results in the sum of the integrals.
5) Integrals of positive functions over a range are positive, and negative if the function is negative.
6) Integrals can be compared based on the relative values of the integrands.
7) Changing the limits of integration flips the sign of the integral.
8) Integrals of odd functions over a symmetric range are zero, and integrals of even functions are twice the integral over
Slideshare is discontinuing its Slidecast feature as of February 28, 2014. Existing Slidecasts will be converted to static presentations without audio by April 30, 2014. The document informs users that new slidecasts can be found on myPlick.com or the author's blog starting in 2014. However, myPlick proved unreliable, so future slidecasts will instead be hosted on the author's YouTube channel.
The document discusses different methods for factorising expressions:
1) Looking for a common factor and dividing it out of all terms
2) Using the difference of two squares formula (a2 - b2 = (a - b)(a + b))
3) Factorising quadratic trinomials into two binomial factors by identifying the values that multiply to give the constant term and sum to give the coefficient of the linear term.
The document provides information on index laws and the meaning of indices in algebra:
- Index laws state that am × an = am+n, am ÷ an = am-n, and (am)n = amn. Exponents can be added or subtracted when multiplying or dividing terms with the same base.
- Positive exponents indicate a term is raised to a power. Negative exponents indicate a root is being taken. Terms with exponents are evaluated from left to right.
- Examples demonstrate how to simplify expressions using index laws and interpret different types of indices.
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
The document discusses integrating derivatives of functions. It states that the integral of the derivative of a function f(x) is equal to the natural log of f(x) plus a constant. It then provides examples of integrating several derivatives: (i) ∫(1/(7-3x)) dx = -1/3 log(7-3x) + c, (ii) ∫(1/(8x+5)) dx = 1/8 log(8x+5) + c, and (iii) ∫(x5/(x-2)) dx = 1/6 log(x6-2) + c. It also discusses techniques for integrating fractions by polynomial long division and finds
The document discusses logarithms and their properties. Logarithms are defined as the inverse of exponentials. If y = ax, then x = loga y. The natural logarithm is log base e, written as ln. Properties of logarithms include: loga m + loga n = loga mn; loga m - loga n = loga(m/n); loga mn = n loga m; loga 1 = 0; loga a = 1. Examples of evaluating logarithmic expressions are provided.
The document discusses relationships between the coefficients and roots of polynomials. It states that for a polynomial P(x) = axn + bxn-1 + cxn-2 + ..., the sum of the roots equals -b/a, the sum of the roots taken two at a time equals c/a, and so on for higher order terms. It also provides examples of using these relationships to find the sums of roots for a given polynomial.
P
4
3
2
The document discusses properties of polynomials with multiple roots. It first proves that if a polynomial P(x) has a root x = a of multiplicity m, then the derivative of P(x), P'(x), will have a root x = a of multiplicity m-1. It then provides an example of solving a cubic equation given it has a double root. Finally, it examines a quartic polynomial and shows that its root α cannot be 0, 1, or -1, and that 1/
The document discusses factorizing complex expressions. The main points are:
- If a polynomial's coefficients are real, its roots will appear in complex conjugate pairs.
- Any polynomial of degree n can be factorized into a mixture of quadratic and linear factors over real numbers, or into n linear factors over complex numbers.
- Odd degree polynomials must have at least one real root.
- Examples of factorizing polynomials over both real and complex numbers are provided.
The document describes the Trapezoidal Rule for approximating the area under a curve between two points. It shows that the area A is estimated by dividing the region into trapezoids with height equal to the function values at the interval endpoints and bases equal to the intervals. In general, the area is approximated as the sum of the areas of each trapezoid, which is equal to the average of the endpoint function values multiplied by the interval length.
The document discusses methods for calculating the volumes of solids of revolution. It provides formulas for finding volumes when an area is revolved around either the x-axis or y-axis. Examples are given for finding volumes of common solids like cones, spheres, and others. Steps are shown for using the formulas to calculate volumes based on given functions and limits of revolution.
The document discusses different methods for calculating the area under a curve or between curves.
(1) The area below the x-axis is given by the integral of the function between the bounds, which can be positive or negative depending on whether the area is above or below the x-axis.
(2) To calculate the area on the y-axis, the function is solved for x in terms of y, then the bounds are substituted into the integral of this new function with respect to y.
(3) The area between two curves is calculated by taking the integral of the upper curve minus the integral of the lower curve, both between the same bounds on the x-axis.
The document discusses 8 properties of definite integrals:
1) Integrating polynomials results in a fraction.
2) Constants can be factored out of integrals.
3) Integrals of sums are equal to the sum of integrals.
4) Splitting an integral range results in the sum of the integrals.
5) Integrals of positive functions over a range are positive, and negative if the function is negative.
6) Integrals can be compared based on the relative values of the integrands.
7) Changing the limits of integration flips the sign of the integral.
8) Integrals of odd functions over a symmetric range are zero, and integrals of even functions are twice the integral over
How to Make a Field Mandatory in Odoo 17Celine George
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptxEduSkills OECD
Iván Bornacelly, Policy Analyst at the OECD Centre for Skills, OECD, presents at the webinar 'Tackling job market gaps with a skills-first approach' on 12 June 2024
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.
Reimagining Your Library Space: How to Increase the Vibes in Your Library No ...Diana Rendina
Librarians are leading the way in creating future-ready citizens – now we need to update our spaces to match. In this session, attendees will get inspiration for transforming their library spaces. You’ll learn how to survey students and patrons, create a focus group, and use design thinking to brainstorm ideas for your space. We’ll discuss budget friendly ways to change your space as well as how to find funding. No matter where you’re at, you’ll find ideas for reimagining your space in this session.
2. Maxima & Minima
Problems
When solving word problems we need to;
3. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
4. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising
5. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising
b) one will be some information given in the problem
6. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising
b) one will be some information given in the problem
(2) Rewrite the equation we are trying to minimise/maximise with
only one variable
7. Maxima & Minima
Problems
When solving word problems we need to;
(1) Reduce the problem to a set of equations
(there will usually be two)
a) one will be what you are maximising/minimising
b) one will be some information given in the problem
(2) Rewrite the equation we are trying to minimise/maximise with
only one variable
(3) Use calculus to solve the problem
8. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
9. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
10. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s
s
11. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
12. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A s 2 rs 1
13. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A s 2 rs 1 6 s 2r 362
14. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A s 2 rs 1 6 s 2r 362
make r the subject in 2
15. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A s 2 rs 1 6 s 2r 362
make r the subject in 2
2r 36 6 s
r 18 3s
16. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A s 2 rs 1 6 s 2r 362
make r the subject in 2
2r 36 6 s
r 18 3s
substitute into 1
17. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A s 2 rs 1 6 s 2r 362
make r the subject in 2
2r 36 6 s
r 18 3s
substitute into 1
A s 2 18 3s s
18. e.g. A rope 36 metres in length is cut into two pieces. The first is bent
to form a square, the other forms a rectangle with one side the
same length as the square.
Find the length of the square if the sum of the areas is to be a
maximum.
s r
s s
A s 2 rs 1 6 s 2r 362
make r the subject in 2
2r 36 6 s
r 18 3s
substitute into 1
A s 2 18 3s s
A s 2 18s 3s 2
A 18s 2 s 2
21. A 18s 2 s 2
dA
18 4 s
ds
d2A
2
4
ds
22. A 18s 2 s 2
dA
18 4 s
ds
d2A
2
4
ds
dA
Stationary points occur when 0
ds
23. A 18s 2 s 2
dA
18 4 s
ds
d2A
2
4
ds
dA
Stationary points occur when 0
ds
i.e.18 4 s 0
9
s
2
24. A 18s 2 s 2
dA
18 4 s
ds
d2A
2
4
ds
dA
Stationary points occur when 0
ds
i.e.18 4 s 0
9
s
2
9 d2A
when s , 2 4 0
2 ds
25. A 18s 2 s 2
dA
18 4 s
ds
d2A
2
4
ds
dA
Stationary points occur when 0
ds
i.e.18 4 s 0
9
s
2
9 d2A
when s , 2 4 0
2 ds
1
when the side length of the square is 4 metres, the area is a maximum
2
26. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
27. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r
h
28. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V r 2 h1
h
29. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V r 2 h1
12 r 2 2rh 2
h
30. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V r 2 h1
12 r 2 2rh 2
make h the subject in 2
h
31. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V r 2 h1
12 r 2 2rh 2
make h the subject in 2
h
2rh 12 r 2
12 r 2
h
2r
32. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V r 2 h1
12 r 2 2rh 2
make h the subject in 2
h
2rh 12 r 2
12 r 2
h
2r
substitute into 1
33. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V r 2 h1
12 r 2 2rh 2
make h the subject in 2
h
2rh 12 r 2
12 r 2
h
2r
substitute into 1
12 r 2
V r 2
2r
34. (ii) An open cylindrical bucket is to made so as to hold the largest
possible volume of liquid.
If the total surface area of the bucket is 12 units 2, find the height,
radius and volume.
r V r 2 h1
12 r 2 2rh 2
make h the subject in 2
h
2rh 12 r 2
12 r 2
h
2r
substitute into 1
12 r 2
V r 2
2r
1
V 6r r 3
2
37. 1
V 6r r 3
2
dV 3
6 r 2
dr 2
d 2V
2
3r
dr
38. 1
V 6r r 3 Stationary points occur when
dV
0
2 dr
dV 3
6 r 2
dr 2
d 2V
2
3r
dr
39. 1
V 6r r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
6 r 2 i.e. 6 r 0
dr 2 2
d 2V 3 2
2
3r r 6
dr 2
r2 4
r 2
40. 1
V 6r r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
6 r 2 i.e. 6 r 0
dr 2 2
d 2V 3 2
2
3r r 6
dr 2
r2 4
d 2V r 2
when r 2, 2 6 0
dr
41. 1
V 6r r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
6 r 2 i.e. 6 r 0
dr 2 2
d 2V 3 2
2
3r r 6
dr 2
r2 4
d 2V r 2
when r 2, 2 6 0
dr
when r 2, V is a maximum
42. 1
V 6r r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
6 r 2 i.e. 6 r 0
dr 2 2
d 2V 3 2
2
3r r 6
dr 2
r2 4
d 2V r 2
when r 2, 2 6 0
dr
when r 2, V is a maximum
12 2 2
when r 2, h
22
2
43. 1
V 6r r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
6 r 2 i.e. 6 r 0
dr 2 2
d 2V 3 2
2
3r r 6
dr 2
r2 4
d 2V r 2
when r 2, 2 6 0
dr
when r 2, V is a maximum
12 2 2
when r 2, h V 2 2
2
22
8
2
44. 1
V 6r r 3 Stationary points occur when
dV
0
2 dr
dV 3 3 2
6 r 2 i.e. 6 r 0
dr 2 2
d 2V 3 2
2
3r r 6
dr 2
r2 4
d 2V r 2
when r 2, 2 6 0
dr
when r 2, V is a maximum
12 2 2
when r 2, h V 2 2
2
22
8
2
maximum volume is 8 units 3 when the radius and height are both 2 units