When a quantity changes in proportion to itself (for instance, bacteria reproduction or radioactive decay), the growth or decay is exponential in nature. There are many many examples of this to be found.
When a quantity changes in proportion to itself (for instance, bacteria reproduction or radioactive decay), the growth or decay is exponential in nature. There are many many examples of this to be found.
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After immersing yourself in the blue book and its red counterpart, attending DDD-focused conferences, and applying tactical patterns, you're left with a crucial question: How do I ensure my design is effective? Tactical patterns within Domain-Driven Design (DDD) serve as guiding principles for creating clear and manageable domain models. However, achieving success with these patterns requires additional guidance. Interestingly, we've observed that a set of constraints initially designed for training purposes remarkably aligns with effective pattern implementation, offering a more ‘mechanical’ approach. Let's explore together how Object Calisthenics can elevate the design of your tactical DDD patterns, offering concrete help for those venturing into DDD for the first time!
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Welcome to UiPath Test Automation using UiPath Test Suite series part 3. In this session, we will cover desktop automation along with UI automation.
Topics covered:
UI automation Introduction,
UI automation Sample
Desktop automation flow
Pradeep Chinnala, Senior Consultant Automation Developer @WonderBotz and UiPath MVP
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
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All these questions and more will be explored as we talk about matching clients’ needs with what your agency offers without pulling teeth or pulling your hair out. Practical tips, and strategies for successful relationship building that leads to closing the deal.
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.
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Sidekick Solutions uses Bonterra Impact Management (fka Social Solutions Apricot) and automation solutions to integrate data for business workflows.
We believe integration and automation are essential to user experience and the promise of efficient work through technology. Automation is the critical ingredient to realizing that full vision. We develop integration products and services for Bonterra Case Management software to support the deployment of automations for a variety of use cases.
This video focuses on the notifications, alerts, and approval requests using Slack for Bonterra Impact Management. The solutions covered in this webinar can also be deployed for Microsoft Teams.
Interested in deploying notification automations for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
Essentials of Automations: Optimizing FME Workflows with ParametersSafe Software
Are you looking to streamline your workflows and boost your projects’ efficiency? Do you find yourself searching for ways to add flexibility and control over your FME workflows? If so, you’re in the right place.
Join us for an insightful dive into the world of FME parameters, a critical element in optimizing workflow efficiency. This webinar marks the beginning of our three-part “Essentials of Automation” series. This first webinar is designed to equip you with the knowledge and skills to utilize parameters effectively: enhancing the flexibility, maintainability, and user control of your FME projects.
Here’s what you’ll gain:
- Essentials of FME Parameters: Understand the pivotal role of parameters, including Reader/Writer, Transformer, User, and FME Flow categories. Discover how they are the key to unlocking automation and optimization within your workflows.
- Practical Applications in FME Form: Delve into key user parameter types including choice, connections, and file URLs. Allow users to control how a workflow runs, making your workflows more reusable. Learn to import values and deliver the best user experience for your workflows while enhancing accuracy.
- Optimization Strategies in FME Flow: Explore the creation and strategic deployment of parameters in FME Flow, including the use of deployment and geometry parameters, to maximize workflow efficiency.
- Pro Tips for Success: Gain insights on parameterizing connections and leveraging new features like Conditional Visibility for clarity and simplicity.
We’ll wrap up with a glimpse into future webinars, followed by a Q&A session to address your specific questions surrounding this topic.
Don’t miss this opportunity to elevate your FME expertise and drive your projects to new heights of efficiency.
Dev Dives: Train smarter, not harder – active learning and UiPath LLMs for do...UiPathCommunity
💥 Speed, accuracy, and scaling – discover the superpowers of GenAI in action with UiPath Document Understanding and Communications Mining™:
See how to accelerate model training and optimize model performance with active learning
Learn about the latest enhancements to out-of-the-box document processing – with little to no training required
Get an exclusive demo of the new family of UiPath LLMs – GenAI models specialized for processing different types of documents and messages
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Speakers:
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👩🏫 Lenka Dulovicova, Product Program Manager, UiPath
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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.
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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
3. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane,
z = f(x, y)
y g(x, y) = 0
x
4. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
z = f(x, y)
Domain
g(x, y) = 0
y x
5. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
The set of points (x, y, z = f(x, y))
z = f(x, y)
with (x, y) on the curve form a
"trail" on the surface defined by
f(x, y).
Domain
g(x, y) = 0
x y
6. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
The set of points (x, y, z = f(x, y))
z = f(x, y)
with (x, y) on the curve form a
"trail" on the surface defined by
f(x, y). Where are the extrema on
this trail? That is, where are the
highest point and the lowest point
on this trail?
Domain
g(x, y) = 0
x y
7. Lagrange Multipliers
Let z = f(x,y) be a function in x and y and g(x, y) = 0 be an
equation in x and y. The graph of z = f(x, y) is a surface and
g(x, y) = 0 is a curve in the xy-plane, and we assume this
curve is in the domain of f(x, y).
The set of points (x, y, z = f(x, y))
z = f(x, y)
with (x, y) on the curve form a
"trail" on the surface defined by
f(x, y). Where are the extrema on
this trail? That is, where are the
highest point and the lowest point
on this trail? The equation
g(x, y) = 0 is called the
Domain constraint equation in the set up.
g(x, y) = 0
x These types of problems are
y
known as the "extrema with
constraints" problems.
9. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
Level curves z = c
10. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
Level curves z = c
11. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
Level curves z = c
y
x
Level curves z = c
12. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Level curves z = c
y
x
Level curves z = c
13. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Thus at an extremum on the trail
over g(x, y) = 0, its tangent must be Level curves z = c
g(x, y) = 0
parallel to the tangent to the contour
y
lines. x
Level curves z = c
14. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Thus at an extremum on the trail
over g(x, y) = 0, its tangent must be Level curves z = c
g(x, y) = 0
parallel to the tangent to the contour
y
lines. Equivalently, at the extremum, x
the normal vector to the constraint
N = ∇g(x, y), and the gradient
to the surface ∇ f(x, y), must be
parallel,
Level curves z = c
15. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines
z = f(x, y)
of z = f(x, y) in the xy plane,
at a point P where the tangent to
g(x, y) = 0 is different from the tangent
to the contour line means the trail
cuts across the contour line, thus
P
the trail is still changing altitude at P.
So P could not be an extremum.
Thus at an extremum on the trail
over g(x, y) = 0, its tangent must be Level curves z = c
g(x, y) = 0
parallel to the tangent to the contour
y
lines. Equivalently, at the extremum, x
the normal vector to the constraint
N = ∇g(x, y), and the gradient
to the surface ∇ f(x, y), must be
parallel, i.e. there is a λ such that:
Level curves z = c
∇g(x, y) = λ* ∇ f(x, y).
16. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines of z = f(x, y) in the xy
plane. At a point P where the tangent to g(x, y) = 0 is
different from the tangent to the contour line means the trail
cuts across the contour line at P, thus it is still changing
altitude. Therefore P could not be an extremum. Thus the
extrema points on g(x, y) = 0 must be tangential to the
contour lines. Equivalently, at the extremum, the gradient to
g(x, y) = 0 and the gradient to the contour line must be the
multiple of each other.
P
g(x, y) = 0
17. Lagrange Multipliers
Plot g(x, y) = 0 over the contour lines of z = f(x, y) in the xy
plane. At a point P where the tangent to g(x, y) = 0 is
different from the tangent to the contour line means the trail
cuts across the contour line at P, thus it is still changing
altitude. Therefore P could not be an extremum. Thus the
extrema points on g(x, y) = 0 must be tangential to the
contour lines. Equivalently, at the extremum, the gradient to
g(x, y) = 0 and the gradient to the contour line must be the
multiple of each other, that is, there is a λ such that:
∇g(x, y) = λ* ∇ f(x, y) or that
P
dg i + dg j = λ ( df i + df j )
dx dy dx dy
N= ∇g(x, y)
so we have gx = λfx
∇f(x, y) the system: gy = λfy
g(x, y) = 0
g(x, y) = 0 We solve this syetem for the extrema.
20. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
21. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx
gy = λfy
g(x, y) = 0
22. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
23. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x
From eq1, we get λ = y ,
24. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
25. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
26. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2
27. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
28. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2
29. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
30. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
31. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
Substitute y = -x into eq3, we get 2x2 = 2
32. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
Substitute y = -x into eq3, we get 2x2 = 2 x = ±1
33. Lagrange Multipliers
The variable λ is called a multiplier. In more complex problems,
we might use many differernt multipliers.
Example: Find the extrema of f(x, y) = xy given the constraint
g(x, y) = x2 + y2 – 2 = 0.
We are to solve: gx = λfx that is: 2x = λy eq1
gy = λfy 2y = λx eq2
g(x, y) = 0 x2 + y2 = 2 eq3
2x 2y
From eq1, we get λ = y , from eq2, we get λ = x .
2y 2x
Set them equal: x = y
we get 2y2 = 2x2 y = ±x
Substitute y = x into eq3, we get 2x2 = 2 x = ±1
If y = x, there are two pairs of solutions (1, 1), (-1, -1).
Substitute y = -x into eq3, we get 2x2 = 2 x = ±1
If y = -x, this gives two more pairs of solutions (1, -1), (-1, 1).
35. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y) = xy:
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
36. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y) = xy:
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
So (1, 1) , (-1, -1) are maxima,
and (-1, 1), (1, -1) are minima.
37. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y):
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
So (1, 1) , (-1, -1) are maxima,
and (-1, 1), (1, -1) are minima.
38. Lagrange Multipliers
To find out which is a max, a min,
or neither, evaluate each with f(x, y):
f(1, 1) = 1, f(-1, -1) =1,
f(-1, 1) = -1, f(1, -1) = -1,
So (1, 1) , (-1, -1) are maxima,
and (-1, 1), (1, -1) are minima.
The procedure above is called the method of "Lagrange
Multipliers". It can be used in the case of finding extrema of a
three varaible function w=f(x, y, z) with a constraint
g(x, y, z) = 0. In the three variabe case, we solve the systems:
gx = λfx
gy = λfy
gz = λfz
g(x, y, z) = 0
41. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
42. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0.
43. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx
gy = λfy
gz = λfz
g(x, y, z) = 0
44. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy
gz = λfz
g(x, y, z) = 0
45. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz
g(x, y, z) = 0
46. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
47. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz
48. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz y = λyz – z
49. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz y = λyz – z
y = (λy – 1)z
50. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z:
y + z = λyz y = λyz – z
y = (λy – 1)z
y
=z
λy – 1
51. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy
y = (λy – 1)z
y
=z
λy – 1
52. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z
y
=z
λy – 1
53. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z y = (λy – 1)x
y
=z
λy – 1
54. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z y = (λy – 1)x
y y
=z =x
λy – 1 λy – 1
55. Lagrange Multipliers
Example: Find the largest volume possibe of a box with a
fixed total surface area.
The volume is V = xyz.
The constraint is 2xy + 2xz + 2yz = c or xy + xz + yz = c/2
Hence g(x, y, z) = xy + xz + yz – c/2 = 0. Set up the system:
gx = λfx y + z = λyz eq1
gy = λfy x + z = λxz eq2
gz = λfz x + y = λxy eq3
xy + xz + yz = c/2 eq4
g(x, y, z) = 0
From eq1, we solve for z: From eq3, we solve for x:
y + z = λyz y = λyz – z x + y = λxy y = λxy – x
y = (λy – 1)z y = (λy – 1)x
y y
=z =x
λy – 1 λy – 1
Hence either x = z or λy – 1 = 0.
58. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
59. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
60. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
61. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
The system is symmetric so similar algebra yields x = y
and y = z.
62. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
The system is symmetric so similar algebra yields x = y
and y = z.
Therefore x = y = z and the box must be a cube.
63. Lagrange Multipliers
Assume λy – 1 = 0, then from y = (λy – 1)z, we get y = 0.
Put this in eq1 and eq3, then x and z must be 0 also,
which is impossibe because of eq4 .
Therefore x = z.
The system is symmetric so similar algebra yields x = y
and y = z.
Therefore x = y = z and the box must be a cube.