In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary). Partial derivatives are used in vector calculus and differential geometry.
A very wide spectrum of optimization problems can be efficiently solved with proximal gradient methods which hinge on the celebrated forward-backward splitting (FBS) schema. But such first-order methods are only effective when low or medium accuracy is required and are known to be rather slow or even impractical for badly conditioned problems. Moreover, the straightforward introduction of second-order (Hessian) information is beset with shortcomings as, typically, at every iteration we need to solve a non-separable optimisation problem. In this talk we will follow a different route to the solution of such optimisation problems. We will recast non-smooth optimisation problems as the minimisation of a real-valued, continuously differentiable function known as the forward-backward envelope. We will then employ a semismooth Newton method to solve the equivalent optimisation problem instead of the original one. We will then apply the proposed semismooth Newton method to L1-regularised least squares (LASSO) problems which is motivated by an an interesting application: recursive compressed sensing. Compressed sensing is a signal processing methodology for the reconstruction of sparsely sampled signals and it offers a new paradigm for sampling signals based on their innovation, that is, the minimum number of coefficients sufficient to accurately represent it in an appropriately selected basis. Compressed sensing leads to a lower sampling rate compared to theories using some fixed basis and has many applications in image processing, medical imaging and MRI, photography, holography, facial recognition, radio astronomy, radar technology and more. The traditional compressed sensing approach is naturally offline, in that it amounts to sparsely sampling and reconstructing a given dataset. Recently, an online algorithm for performing compressed sensing on streaming data was proposed; the scheme uses recursive sampling of the input stream and recursive decompression to accurately estimate stream entries from the acquired noisy measurements. We will see how we can tailor the forward-backward Newton method to solve recursive compressed sensing problems at one tenth of the time required by other algorithms such as ISTA, FISTA, ADMM and interior-point methods (L1LS).
Properties of Functions
Odd and Even Functions
Periodic Functions
Monotonic Functions
Bounded Functions
Maxima and Minima of Functions
Inverse Function
Sequence and Series
Software Delivery At the Speed of AI: Inflectra Invests In AI-Powered QualityInflectra
In this insightful webinar, Inflectra explores how artificial intelligence (AI) is transforming software development and testing. Discover how AI-powered tools are revolutionizing every stage of the software development lifecycle (SDLC), from design and prototyping to testing, deployment, and monitoring.
Learn about:
• The Future of Testing: How AI is shifting testing towards verification, analysis, and higher-level skills, while reducing repetitive tasks.
• Test Automation: How AI-powered test case generation, optimization, and self-healing tests are making testing more efficient and effective.
• Visual Testing: Explore the emerging capabilities of AI in visual testing and how it's set to revolutionize UI verification.
• Inflectra's AI Solutions: See demonstrations of Inflectra's cutting-edge AI tools like the ChatGPT plugin and Azure Open AI platform, designed to streamline your testing process.
Whether you're a developer, tester, or QA professional, this webinar will give you valuable insights into how AI is shaping the future of software delivery.
Kubernetes & AI - Beauty and the Beast !?! @KCD Istanbul 2024Tobias Schneck
As AI technology is pushing into IT I was wondering myself, as an “infrastructure container kubernetes guy”, how get this fancy AI technology get managed from an infrastructure operational view? Is it possible to apply our lovely cloud native principals as well? What benefit’s both technologies could bring to each other?
Let me take this questions and provide you a short journey through existing deployment models and use cases for AI software. On practical examples, we discuss what cloud/on-premise strategy we may need for applying it to our own infrastructure to get it to work from an enterprise perspective. I want to give an overview about infrastructure requirements and technologies, what could be beneficial or limiting your AI use cases in an enterprise environment. An interactive Demo will give you some insides, what approaches I got already working for real.
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
"Impact of front-end architecture on development cost", Viktor TurskyiFwdays
I have heard many times that architecture is not important for the front-end. Also, many times I have seen how developers implement features on the front-end just following the standard rules for a framework and think that this is enough to successfully launch the project, and then the project fails. How to prevent this and what approach to choose? I have launched dozens of complex projects and during the talk we will analyze which approaches have worked for me and which have not.
Accelerate your Kubernetes clusters with Varnish CachingThijs Feryn
A presentation about the usage and availability of Varnish on Kubernetes. This talk explores the capabilities of Varnish caching and shows how to use the Varnish Helm chart to deploy it to Kubernetes.
This presentation was delivered at K8SUG Singapore. See https://feryn.eu/presentations/accelerate-your-kubernetes-clusters-with-varnish-caching-k8sug-singapore-28-2024 for more details.
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf91mobiles
91mobiles recently conducted a Smart TV Buyer Insights Survey in which we asked over 3,000 respondents about the TV they own, aspects they look at on a new TV, and their TV buying preferences.
Key Trends Shaping the Future of Infrastructure.pdfCheryl Hung
Keynote at DIGIT West Expo, Glasgow on 29 May 2024.
Cheryl Hung, ochery.com
Sr Director, Infrastructure Ecosystem, Arm.
The key trends across hardware, cloud and open-source; exploring how these areas are likely to mature and develop over the short and long-term, and then considering how organisations can position themselves to adapt and thrive.
Search and Society: Reimagining Information Access for Radical FuturesBhaskar Mitra
The field of Information retrieval (IR) is currently undergoing a transformative shift, at least partly due to the emerging applications of generative AI to information access. In this talk, we will deliberate on the sociotechnical implications of generative AI for information access. We will argue that there is both a critical necessity and an exciting opportunity for the IR community to re-center our research agendas on societal needs while dismantling the artificial separation between the work on fairness, accountability, transparency, and ethics in IR and the rest of IR research. Instead of adopting a reactionary strategy of trying to mitigate potential social harms from emerging technologies, the community should aim to proactively set the research agenda for the kinds of systems we should build inspired by diverse explicitly stated sociotechnical imaginaries. The sociotechnical imaginaries that underpin the design and development of information access technologies needs to be explicitly articulated, and we need to develop theories of change in context of these diverse perspectives. Our guiding future imaginaries must be informed by other academic fields, such as democratic theory and critical theory, and should be co-developed with social science scholars, legal scholars, civil rights and social justice activists, and artists, among others.
PHP Frameworks: I want to break free (IPC Berlin 2024)Ralf Eggert
In this presentation, we examine the challenges and limitations of relying too heavily on PHP frameworks in web development. We discuss the history of PHP and its frameworks to understand how this dependence has evolved. The focus will be on providing concrete tips and strategies to reduce reliance on these frameworks, based on real-world examples and practical considerations. The goal is to equip developers with the skills and knowledge to create more flexible and future-proof web applications. We'll explore the importance of maintaining autonomy in a rapidly changing tech landscape and how to make informed decisions in PHP development.
This talk is aimed at encouraging a more independent approach to using PHP frameworks, moving towards a more flexible and future-proof approach to PHP development.
UiPath Test Automation using UiPath Test Suite series, part 4DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
What will you get from this session?
1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
Topics covered:
Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
2. Applications in Optimization
In this section, we solve various extrema problems
in real world applications. These problems are
referred to as “optimization problems” in
mathematics, and derivatives play a major role in
such problems.
3. Applications in Optimization
In this section, we solve various extrema problems
in real world applications. These problems are
referred to as “optimization problems” in
mathematics, and derivatives play a major role in
such problems. We begin with an
extrema-existence–problem.
4. Applications in Optimization
In this section, we solve various extrema problems
in real world applications. These problems are
referred to as “optimization problems” in
mathematics, and derivatives play a major role in
such problems. We begin with an
extrema-existence–problem.
Let y be a function defined over an
interval V, y may or may not have
any extremum in V.
5. Applications in Optimization
In this section, we solve various extrema problems
in real world applications. These problems are
referred to as “optimization problems” in
mathematics, and derivatives play a major role in
such problems. We begin with an
extrema-existence–problem.
Let y be a function defined over an
interval V, y may or may not have
any extremum in V. Here are two
examples of well behaved functions
that don’t have extrema.
6. Applications in Optimization
In this section, we solve various extrema problems
in real world applications. These problems are
referred to as “optimization problems” in
mathematics, and derivatives play a major role in
such problems. We begin with an
extrema-existence–problem.
Let y be a function defined over an
interval V, y may or may not have
any extremum in V. Here are two
examples of well behaved functions
that don’t have extrema.
Let y = tan(x) over the interval
V = (–π/2, π/2) as shown.
y=tan(x)
x
y
–π/2 π/2
7. Applications in Optimization
In this section, we solve various extrema problems
in real world applications. These problems are
referred to as “optimization problems” in
mathematics, and derivatives play a major role in
such problems. We begin with an
extrema-existence–problem.
Let y be a function defined over an
interval V, y may or may not have
any extremum in V. Here are two
examples of well behaved functions
that don’t have extrema.
Let y = tan(x) over the interval
V = (–π/2, π/2) as shown. There is no
extremum in V because V is open.
y=tan(x)
x
y
–π/2 π/2
8. Applications in Optimization
Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.
9. Applications in Optimization
Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.
x
y=g(x)
For example, let y = g(x) be the
function over the closed interval
U = [–π/2, π/2] as shown.
y
(0, 1)
–π/2 π/2
(0, –1)
10. Applications in Optimization
Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.
x
y=g(x)
For example, let y = g(x) be the
function over the closed interval
U = [–π/2, π/2] as shown. Note that
even though g(x) is bounded between
y
(0, 1)
–π/2 π/2
(0, –1)
1 and –1, it still doesn’t have any extremum in U.
11. Applications in Optimization
Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.
x
y=g(x)
For example, let y = g(x) be the
function over the closed interval
U = [–π/2, π/2] as shown. Note that
even though g(x) is bounded between
y
(0, 1)
–π/2 π/2
(0, –1)
1 and –1, it still doesn’t have any extremum in U.
(Extrema Theorem for Continuous Functions)
12. Applications in Optimization
Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.
x
y=g(x)
For example, let y = g(x) be the
function over the closed interval
U = [–π/2, π/2] as shown. Note that
even though g(x) is bounded between
y
(0, 1)
–π/2 π/2
(0, –1)
1 and –1, it still doesn’t have any extremum in U.
(Extrema Theorem for Continuous Functions)
Let y = f(x) be a continuous function defined over a
closed interval V = [a, b], then both the absolute max.
and the absolute min. exist in V.
13. Applications in Optimization
Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.
x
y=g(x)
For example, let y = g(x) be the
function over the closed interval
U = [–π/2, π/2] as shown. Note that
even though g(x) is bounded between
y
(0, 1)
–π/2 π/2
(0, –1)
1 and –1, it still doesn’t have any extremum in U.
(Extrema Theorem for Continuous Functions)
Let y = f(x) be a continuous function defined over a
closed interval V = [a, b], then both the absolute max.
and the absolute min. exist in V. Furthermore, the
absolute extrema must occur where f '(x) = 0,
14. Applications in Optimization
Even if we close the interval V,
it’s still not enough to guarantee
the existence of an extremum.
x
y=g(x)
For example, let y = g(x) be the
function over the closed interval
U = [–π/2, π/2] as shown. Note that
even though g(x) is bounded between
y
(0, 1)
–π/2 π/2
(0, –1)
1 and –1, it still doesn’t have any extremum in U.
(Extrema Theorem for Continuous Functions)
Let y = f(x) be a continuous function defined over a
closed interval V = [a, b], then both the absolute max.
and the absolute min. exist in V. Furthermore, the
absolute extrema must occur where f '(x) = 0, or where
f'(x) is UDF, or they occur at the end points {a, b}.
17. Applications in Optimization
Here are examples of each type of extrema.
A.
a
x
b
In A. The absolute maximum
occurs at x = c where
f '(c) = 0, and the absolute
minimum occurs at the end
point x = b.
c
A f'=0 ab. max.
An end–point
ab. min.
18. Applications in Optimization
Here are examples of each type of extrema.
A.
a
An end–point
ab. min.
x
A f'=0 ab. max.
b
B.
a b x
In A. The absolute maximum
occurs at x = c where
f '(c) = 0, and the absolute
minimum occurs at the end
point x = b.
c
c d
19. Applications in Optimization
Here are examples of each type of extrema.
A.
a
x
b
B.
a b x
In A. The absolute maximum
occurs at x = c where
f '(c) = 0, and the absolute
minimum occurs at the end
point x = b.
c
c d
In B. The absolute maximum
occurs at x = c where
f '(c) is UDF, and the absolute
minimum occurs at x = d
where f '(d) = 0.
A f'=0 ab. max.
An end–point
ab. min.
A UDF– f' ab. max.
A f'=0
ab. min.
20. Applications in Optimization
Here are examples of each type of extrema.
A.
a
x
b
B.
a b x
In A. The absolute maximum
occurs at x = c where
f '(c) = 0, and the absolute
minimum occurs at the end
point x = b.
c
c d
In B. The absolute maximum
occurs at x = c where
f '(c) is UDF, and the absolute
minimum occurs at x = d
where f '(d) = 0.
A f'=0 ab. max.
An end–point
ab. min.
A UDF– f' ab. max.
A f'=0
ab. min.
Hence we have to consider
points of all three cases when
solving for extrema.
22. Applications in Optimization
Example A. Let y be defined as the following.
x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
23. Applications in Optimization
Example A. Let y be defined as the following.
x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
Find the extrema of y over the
interval [–1 , ½ ].
24. Applications in Optimization
Example A. Let y be defined as the following.
x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
Find the extrema of y over the
interval [–1 , ½ ].
The end–point values are
f(–1) = 1 and f(½) = ¼.
25. Applications in Optimization
Example A. Let y be defined as the following.
x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
Find the extrema of y over the
interval [–1 , ½ ].
x
y
The end–point values are
f(–1) = 1 and f(½) = ¼.
(–1, 1)
–1 ½
(½, ¼ )
26. Applications in Optimization
Example A. Let y be defined as the following.
x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
Find the extrema of y over the
interval [–1 , ½ ].
x
y
The end–point values are
f(–1) = 1 and f(½) = ¼.
y' = 0 at x = 0 so that (0, 0)
is an f'=0–type minimum.
(–1, 1)
–1 ½
(½, ¼ )
27. Applications in Optimization
Example A. Let y be defined as the following.
x2 for 1 > x
y =
2 – x for 1 ≤ x
(1, 1)
x
y
Find the extrema of y over the
interval [–1 , ½ ].
x
y
The end–point values are
f(–1) = 1 and f(½) = ¼.
y' = 0 at x = 0 so that (0, 0)
is an f'=0–type minimum.
Within the interval [–1 , ½ ], there
is no point of the f'–UDF–type.
Hence the ab. max. is at (–1, 1)
and the ab. min is at (0, 0).
(–1, 1)
–1 ½
(½, ¼ )
28. Applications in Optimization
The key here is that if we are to find extrema of a
given continuous function over a closed interval
we have to obtain all three types of points described
above, then select the correct solutions.
29. Applications in Optimization
x
The key here is that if we are to find extrema of a
given continuous function over a closed interval
we have to obtain all three types of points described
above, then select the correct solutions.
For instance, within [–½, 3], y
x = 0 is an f'=0–type solution.
The point (1, 1) at x = 1 is an
f'–UDF–type solution.
(1, 1)
–½ 3
30. Applications in Optimization
x
The key here is that if we are to find extrema of a
given continuous function over a closed interval
we have to obtain all three types of points described
above, then select the correct solutions.
For instance, within [–½, 3], y
x = 0 is an f'=0–type solution.
The point (1, 1) at x = 1 is an
f'–UDF–type solution.
The end–point values are
f(–½) = ¼ and f(3) = –1.
(1, 1)
–½ 3
(3, –1)
31. Applications in Optimization
x
The key here is that if we are to find extrema of a
given continuous function over a closed interval
we have to obtain all three types of points described
above, then select the correct solutions.
For instance, within [–½, 3], y
x = 0 is an f'=0–type solution.
The point (1, 1) at x = 1 is an
f'–UDF–type solution.
The end–point values are
f(–½) = ¼ and f(3) = –1.
By inspection, the ab. max. is at
(1, 1) and the ab. min is at (3, –1).
(1, 1)
–½ 3
(3, –1)
32. Applications in Optimization
x
The key here is that if we are to find extrema of a
given continuous function over a closed interval
we have to obtain all three types of points described
above, then select the correct solutions.
For instance, within [–½, 3], y
x = 0 is an f'=0–type solution.
The point (1, 1) at x = 1 is an
f'–UDF–type solution.
The end–point values are
f(–½) = ¼ and f(3) = –1.
By inspection, the ab. max. is at
(1, 1) and the ab. min is at (3, –1).
(1, 1)
–½ 3
(3, –1)
In the applied problems below, make sure all such
relevant points are considered.
34. Applications in Optimization
Distance Problems
The distance formula is a square–root formula whose
radicand is always nonnegative. To find the extrema
of a distance relation in the form of “√u(x)”,
we drop the square–root and look for the extrema of
u(x) instead.
35. Applications in Optimization
Distance Problems
The distance formula is a square–root formula whose
radicand is always nonnegative. To find the extrema
of a distance relation in the form of “√u(x)”,
we drop the square–root and look for the extrema of
u(x) instead.
Example B. Find the coordinate of the point on the
line y = 2x that is closest to the point (1, 0).
36. Applications in Optimization
Distance Problems
The distance formula is a square–root formula whose
radicand is always nonnegative. To find the extrema
of a distance relation in the form of “√u(x)”,
we drop the square–root and look for the extrema of
u(x) instead.
Example B. Find the coordinate of the point on the
line y = 2x that is closest to the point (1, 0).
x
y=2x
(1, 0)
y
D
37. Applications in Optimization
Distance Problems
The distance formula is a square–root formula whose
radicand is always nonnegative. To find the extrema
of a distance relation in the form of “√u(x)”,
we drop the square–root and look for the extrema of
u(x) instead.
Example B. Find the coordinate of the point on the
line y = 2x that is closest to the point (1, 0).
x
y=2x
(1, 0)
y
(x, 2x)
A generic point on the line is (x, 2x)
and its distance to the point (1, 0)
is D = [(x – 1)2 + (2x – 0 )2]½.
D
38. Applications in Optimization
Distance Problems
The distance formula is a square–root formula whose
radicand is always nonnegative. To find the extrema
of a distance relation in the form of “√u(x)”,
we drop the square–root and look for the extrema of
u(x) instead.
Example B. Find the coordinate of the point on the
line y = 2x that is closest to the point (1, 0).
x
y=2x
(1, 0)
y
(x, 2x)
A generic point on the line is (x, 2x)
and its distance to the point (1, 0)
is D = [(x – 1)2 + (2x – 0 )2]½.
We want to find the x value that
gives the minimal D. Note that D is
defined for all numbers.
D
39. Applications in Optimization
x
(1, 0)
(x, 2x)
We minimize the radicand
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
y
D
40. Applications in Optimization
x
(1, 0)
(x, 2x)
We minimize the radicand
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
and the point is (1/5, 2/5).
y
D
41. Applications in Optimization
x
(1, 0)
(x, 2x)
We minimize the radicand
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
and the point is (1/5, 2/5).
From the geometry it’s obvious
(1/5, 2/5) is the closest point to (1, 0).
y
D
42. Applications in Optimization
x
(1, 0)
(x, 2x)
We minimize the radicand
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
and the point is (1/5, 2/5).
From the geometry it’s obvious
(1/5, 2/5) is the closest point to (1, 0).
½
y
D
Similarly if we wish to find the extrema of eu(x) or
In(u(x)), we may find the extrema of u(x) instead.
This is true because like sqrt(x), eu(x) and In(u(x)) are
increasing functions, i.e. if s < t, then es < et,
so the extrema of eu(x) are the same as u(x).
43. Applications in Optimization
x
(1, 0)
(x, 2x)
We minimize the radicand
u = D2 = (x – 1)2 + (2x – 0 )2
= 5x2 – 2x + 1 instead.
Set u' = 10x – 2 = 0. We get x = 1/5,
and the point is (1/5, 2/5).
From the geometry it’s obvious
(1/5, 2/5) is the closest point to (1, 0).
½
y
D
Similarly if we wish to find the extrema of eu(x) or
In(u(x)), we may find the extrema of u(x) instead.
This is true because like sqrt(x), eu(x) and In(u(x)) are
increasing functions, i.e. if s < t, then es < et,
so the extrema of eu(x) are the same as u(x).
In fact, if f(x) is an increasing function, then the
extrema of f(u(x)) are the same as the ones of u(x).
44. Applications in Optimization
(What is the corresponding statement about the
extrema of f(u(x)) and u(x) if f(x) is a decreasing
function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)
45. Applications in Optimization
(What is the corresponding statement about the
extrema of f(u(x)) and u(x) if f(x) is a decreasing
function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)
Area Problems
In area optimization problems, the main task is to
express the area in question in a single well chosen
variable x as A(x).
46. Applications in Optimization
(What is the corresponding statement about the
extrema of f(u(x)) and u(x) if f(x) is a decreasing
function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)
Area Problems
In area optimization problems, the main task is to
express the area in question in a single well chosen
variable x as A(x).
If x represents a specific measurement, as opposed to
a coordinate, then x must be nonnegative (often it’s
also bounded above).
47. Applications in Optimization
(What is the corresponding statement about the
extrema of f(u(x)) and u(x) if f(x) is a decreasing
function? Hint, consider f(x) = –x and f(u(x)) = –u(x).)
Area Problems
In area optimization problems, the main task is to
express the area in question in a single well chosen
variable x as A(x).
If x represents a specific measurement, as opposed to
a coordinate, then x must be nonnegative (often it’s
also bounded above). In such cases consider the
significance when x = 0 (or it’s other boundary value).
Often one of the extrema is at the boundary, and we
are looking for the other extremum.
A drawing is indispensable in any geometric problem.
48. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
to build rectangular regions as
shown to separate the animals.
What is the maximal enclosed
area that is possible? x
x/2
y
49. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
to build rectangular regions as
shown to separate the animals.
What is the maximal enclosed
area that is possible?
The enclosed area is A = xy.
x
x/2
y
50. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
x/2
to build rectangular regions as
y
shown to separate the animals.
What is the maximal enclosed
area that is possible?
x
The enclosed area is A = xy. The total linear length
is 120, that is, 3y + 2½ x = 120
51. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
x/2
to build rectangular regions as
y
shown to separate the animals.
What is the maximal enclosed
area that is possible?
x
The enclosed area is A = xy. The total linear length
is 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6.
52. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
x/2
to build rectangular regions as
y
shown to separate the animals.
What is the maximal enclosed
area that is possible?
x
The enclosed area is A = xy. The total linear length
is 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6.
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.
53. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
x/2
to build rectangular regions as
y
shown to separate the animals.
What is the maximal enclosed
area that is possible?
x
The enclosed area is A = xy. The total linear length
is 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6.
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.
At the boundary values x = 0, 48 (why?) we have the
absolute minimal A = 0.
54. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
x/2
to build rectangular regions as
y
shown to separate the animals.
What is the maximal enclosed
area that is possible?
x
The enclosed area is A = xy. The total linear length
is 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6.
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.
At the boundary values x = 0, 48 (why?) we have the
absolute minimal A = 0. Setting A'(x) = 40 – 5x/3 = 0,
we get x = 24
55. Applications in Optimization
Example C. Farmer Fred wants
to raise chickens, ducks and
pigs. He has 120 yards of fence
x/2
to build rectangular regions as
y
shown to separate the animals.
What is the maximal enclosed
area that is possible?
x
The enclosed area is A = xy. The total linear length
is 120, that is, 3y + 2½ x = 120 or that y = 40 – 5x/6.
So A(x) = x(40 – 5x/6) = 40x– 5x2/6.
At the boundary values x = 0, 48 (why?) we have the
absolute minimal A = 0. Setting A'(x) = 40 – 5x/3 = 0,
we get x = 24 which yields y = 20.
So 20×24 = 480 yd2 must be the maximum area.
56. Applications in Optimization
In fact for any specified partition of the rectangle, the
maximal area always occurs when the total–vertical–
fence–length is equal to the total–horizontal–fence–
length. That is, using half of the fence for vertical
dividers and the other half for the horizontal dividers
will give the maximum enclosed area.
Hence if farmer Joe is to
build an enclosure with 120
yd of fence as shown,
x/2
then the answer is
y
x = 60/3½ = 120/7,
y = 60/3 = 20
with the maximum possible
area of 120/7×20 ≈ 343 yd2. x
57. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
120 yd3
h
r
58. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
120 yd3
h
r
The volume of a cylinder is V = Bh
where B = area of the base
and h = height
59. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
120 yd3
h
r
The volume of a cylinder is V = Bh
where B = area of the base
and h = height so we have120π = πr2h
or h = 120/r2.
60. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
120 yd3
h
r
The volume of a cylinder is V = Bh
where B = area of the base
and h = height so we have120π = πr2h
or h = 120/r2. The surface area of the
cylinder consists of a circular
base and the circular wall.
61. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
The volume of a cylinder is V = Bh
where B = area of the base
120 yd3
and h = height so we have120π = πr2h
or h = 120/r2. The surface area of the
cylinder consists of a circular
base and the circular wall. The base area is πr2.
h
r
62. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
The volume of a cylinder is V = Bh
where B = area of the base
120 yd3
h
and h = height so we have120π = πr2h
or h = 120/r2. The surface area of the
r
cylinder consists of a circular
base and the circular wall. The base area is πr2.
Unroll and flatten the wall, it is an h x 2πr rectangular
sheet.
63. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
The volume of a cylinder is V = Bh
where B = area of the base
120 yd3
h
and h = height so we have120π = πr2h
or h = 120/r2. The surface area of the
r
cylinder consists of a circular
base and the circular wall. The base area is πr2.
Unroll and flatten the wall, it is an h x 2πr rectangular
sheet. Hence the total surface is S = πr2+ 2πrh,
or S = πr2 + 2πr( 1 2 0 )
r2
64. Applications in Optimization
Example D. Farmer Joe wishes to build an open
cylindrical water tank made from sheet–metal that
will hold 120π yd3 of water. What is the least amount
of sheet–metal (in yd2) needed?
The volume of a cylinder is V = Bh
where B = area of the base
120 yd3
h
and h = height so we have120π = πr2h
or h = 120/r2. The surface area of the
r
cylinder consists of a circular
base and the circular wall. The base area is πr2.
Unroll and flatten the wall, it is an h x 2πr rectangular
sheet. Hence the total surface is S = πr2+ 2πrh,
or S = πr2 + 2πr( 1 2 0 ) = πr2 + 240π/r = π(r2 + 240/r)
r2
65. Applications in Optimization
120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
dividing both sides by 2π and clearing
the denominator yields r3 – 120 = 0
66. Applications in Optimization
120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
dividing both sides by 2π and clearing
the denominator yields r3 – 120 = 0
or that r = 1201/3, which gives the
absolute minimal (why?) surface
67. Applications in Optimization
120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
dividing both sides by 2π and clearing
the denominator yields r3 – 120 = 0
or that r = 1201/3, which gives the
absolute minimal (why?) surface
of π(1202/3 + ) = π 360/120 240 1/3
1201/3 yd2.
68. Applications in Optimization
120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
dividing both sides by 2π and clearing
the denominator yields r3 – 120 = 0
or that r = 1201/3, which gives the
absolute minimal (why?) surface
of π(1202/3 + ) = π 360/120 240 1/3
1201/3 yd2.
If the geometric object has a variable angle θ, then θ
may be utilized as the independent variable.
69. Applications in Optimization
120 yd3
h
r
Setting S' = 2π(r – 120/r2) = 0,
dividing both sides by 2π and clearing
the denominator yields r3 – 120 = 0
or that r = 1201/3, which gives the
absolute minimal (why?) surface
of π(1202/3 + ) = π 360/120 240 1/3
1201/3 yd2.
If the geometric object has a variable angle θ, then θ
may be utilized as the independent variable.
Example E.
Find the isosceles triangle
4
having two sides of length
4
4 with the largest area as
shown.
71. Applications in Optimization
There are two observations that
will make the algebra cleaner.
First let’s take advantage of the
symmetry. Draw a perpendicular
line as shown, it cuts the triangle
4
4
into two mirror image right
triangles.
72. Applications in Optimization
There are two observations that
will make the algebra cleaner.
First let’s take advantage of the
symmetry. Draw a perpendicular
line as shown, it cuts the triangle
4
4
into two mirror image right
triangles. We may maximize one
of the right triangles instead,
i.e. to find the largest area
possible of a right triangle with
hypotenuse equal to 4.
4
a
b A
73. Applications in Optimization
There are two observations that
will make the algebra cleaner.
First let’s take advantage of the
symmetry. Draw a perpendicular
line as shown, it cuts the triangle
4
4
into two mirror image right
triangles. We may maximize one
of the right triangles instead,
i.e. to find the largest area
possible of a right triangle with
hypotenuse equal to 4.
4
a
b
A
θ
Second, instead of expressing the area A in terms of
the measurements of the legs a and b,
we express A in terms of the angle θ as shown.
74. Applications in Optimization
A = ½ ab
so in terms of θ we have
4
a
θ
b A
A = ½ 4sin(θ) 4cos(θ)
= 8 sin(θ) cos(θ)
75. Applications in Optimization
A = ½ ab
so in terms of θ we have
Note that 0 ≤ θ ≤ π/2.
4
a
θ
b A
A = ½ 4sin(θ) 4cos(θ)
= 8 sin(θ) cos(θ)
76. Applications in Optimization
A = ½ ab
so in terms of θ we have
Note that 0 ≤ θ ≤ π/2.
Set A'(θ) = 8(cos2(θ) – sin2 (θ)) = 0
4
a
θ
b A
A = ½ 4sin(θ) 4cos(θ)
= 8 sin(θ) cos(θ)
77. Applications in Optimization
A = ½ ab
so in terms of θ we have
Note that 0 ≤ θ ≤ π/2.
Set A'(θ) = 8(cos2(θ) – sin2 (θ)) = 0
4
a
θ
b A
A = ½ 4sin(θ) 4cos(θ)
= 8 sin(θ) cos(θ)
Hence cos(θ) = ±sin (θ)
However, the only solution with 0 ≤ θ ≤ π/2 is θ = π/4
78. Applications in Optimization
A = ½ ab
so in terms of θ we have
Note that 0 ≤ θ ≤ π/2.
Set A'(θ) = 8(cos2(θ) – sin2 (θ)) = 0
4
a
b
Hence cos(θ) = ±sin (θ)
which must be a maximum (why?).
So the largest possible isosceles
triangle in question is the right
isosceles triangle as shown with
area 2A = ½ (4 ×4) = 8.
A
θ
A = ½ 4sin(θ) 4cos(θ)
= 8 sin(θ) cos(θ)
However, the only solution with 0 ≤ θ ≤ π/2 is θ = π/4
4
4 π/2
79. Applications in Optimization
Example F. A duck can walk 3 ft/sec and swim
2 ft/sec at its top speed. It sees a piece of bread
across the pool as shown. What is the path it should
take to get to the bread as soon as possible?
40 ft
pool
30 ft
80. Applications in Optimization
Example F. A duck can walk 3 ft/sec and swim
2 ft/sec at its top speed. It sees a piece of bread
across the pool as shown. What is the path it should
take to get to the bread as soon as possible?
40 ft
pool Select the x as shown in the
picture.
30 ft
x
81. Applications in Optimization
Example F. A duck can walk 3 ft/sec and swim
2 ft/sec at its top speed. It sees a piece of bread
across the pool as shown. What is the path it should
take to get to the bread as soon as possible?
40 ft
pool Select the x as shown in the
picture. (The selection of the
correct x is important. The wrong
choice leads to tangled algebra
as is often the case in these
problems.)
30 ft
x
82. Applications in Optimization
Example F. A duck can walk 3 ft/sec and swim
2 ft/sec at its top speed. It sees a piece of bread
across the pool as shown. What is the path it should
take to get to the bread as soon as possible?
40 ft
pool Select the x as shown in the
picture. (The selection of the
correct x is important. The wrong
choice leads to tangled algebra
as is often the case in these
problems.)
30 ft
Hence if your choice of x
leads to nowhere,
try another choice.
x
83. Applications in Optimization
Example F. A duck can walk 3 ft/sec and swim
2 ft/sec at its top speed. It sees a piece of bread
across the pool as shown. What is the path it should
take to get to the bread as soon as possible?
40 ft
pool
30 ft
x
Select the x as shown in the
picture. (The selection of the
correct x is important. The wrong
choice leads to tangled algebra
as is often the case in these
problems.)
Note that 0 ≤ x ≤ 40, where x = 0
means the duck walks around
the pool and x = 40 means the duck swims all the way.
85. Applications in Optimization
40 ft
pool
30 ft
x
(x2 + 900)½
40 – x
For x ≠ 0, the duck swims a
distance of (x2 + 900)½ and
walks a distance of (40 – x),
86. Applications in Optimization
40 ft
pool
30 ft
x
For x ≠ 0, the duck swims a
distance of (x2 + 900)½ and
walks a distance of (40 – x),
and the total time t it takes is
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
swimming time walking time
87. Applications in Optimization
40 ft
pool
30 ft
x
For x ≠ 0, the duck swims a
distance of (x2 + 900)½ and
walks a distance of (40 – x),
and the total time t it takes is
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
88. Applications in Optimization
40 ft
pool
30 ft
x
For x ≠ 0, the duck swims a
distance of (x2 + 900)½ and
walks a distance of (40 – x),
and the total time t it takes is
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½
89. Applications in Optimization
40 ft
pool
30 ft
x
For x ≠ 0, the duck swims a
distance of (x2 + 900)½ and
walks a distance of (40 – x),
and the total time t it takes is
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½
3x = 2(x2 + 900)½
90. Applications in Optimization
40 ft
pool
30 ft
x
For x ≠ 0, the duck swims a
distance of (x2 + 900)½ and
walks a distance of (40 – x),
and the total time t it takes is
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½
3x = 2(x2 + 900)½
9x2 = 4(x2 + 900)
91. Applications in Optimization
40 ft
pool
30 ft
x
For x ≠ 0, the duck swims a
distance of (x2 + 900)½ and
walks a distance of (40 – x),
and the total time t it takes is
(x2 + 900)½
40 – x
t(x) = (x2 + 900)½
2 +
(40 – x)
3
x –
Set t'(x) = 2(x2 + 900)½
1
3
= 0
We have x = 3 1
2(x2 + 900)½
3x = 2(x2 + 900)½
9x2 = 4(x2 + 900)
5x2 = 3600 x = ±12√5
Discarding the negative answer, x =12√5 ≈ 26.8 ft.
92. Applications in Optimization
When x = 0, the time it takes is not t(0) because t(x)
assume the duck swims at least 30 ft,
93. Applications in Optimization
When x = 0, the time it takes is not t(0) because t(x)
assume the duck swims at least 30 ft, as it would be
the case if the duck is across a river from the bread
instead of a pool.
94. Applications in Optimization
When x = 0, the time it takes is not t(0) because t(x)
assume the duck swims at least 30 ft, as it would be
the case if the duck is across a river from the bread
instead of a pool. Let’s examine the three points.
95. Applications in Optimization
When x = 0, the time it takes is not t(0) because t(x)
assume the duck swims at least 30 ft, as it would be
the case if the duck is across a river from the bread
instead of a pool. Let’s examine the three points.
When x = 0, the duck walks 70 ft and it takes
70/3 = 23 1/3 seconds.
When x = 40 the duck swims all the way and it
takes t(40) = 50/2 = 25 seconds.
At the critical point, t(12√5) ≈ 24.5 seconds.
96. Applications in Optimization
When x = 0, the time it takes is not t(0) because t(x)
assume the duck swims at least 30 ft, as it would be
the case if the duck is across a river from the bread
instead of a pool. Let’s examine the three points.
When x = 0, the duck walks 70 ft and it takes
70/3 = 23 1/3 seconds.
When x = 40 the duck swims all the way and it
takes t(40) = 50/2 = 25 seconds.
At the critical point, t(12√5) ≈ 24.5 seconds.
Therefore the duck should walk around the pool.
97. Applications in Optimization
When x = 0, the time it takes is not t(0) because t(x)
assume the duck swims at least 30 ft, as it would be
the case if the duck is across a river from the bread
instead of a pool. Let’s examine the three points.
When x = 0, the duck walks 70 ft and it takes
70/3 = 23 1/3 seconds.
When x = 40 the duck swims all the way and it
takes t(40) = 50/2 = 25 seconds.
At the critical point, t(12√5) ≈ 24.5 seconds.
Therefore the duck should walk around the pool.
Remarks
1. As noted that if the duck is across a river instead
of a pool, then indeed x = 12√5 is the absolute
minimum because t(0) = swim + walk = 28 1/3 sec.
98. Applications in Optimization
2. Just because the duck walks faster does not
automatically means it should always walk around
the pool. We may think about this by imagining the
duck is slightly faster in walking, say at 2.01 ft/sec,
than swimming at 2 ft/sec. Then of course the duck
should swim toward the bread instead of wasting
time walking around. In fact the closer the walking
speed is to the swimming speed, the more the duck
should swim toward the bread.
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