Maximum and minimum of $x^2+y^2-2x$Farthest and Nearest Point in the ellipse from a focus using...
How did Alan Turing break the enigma code using the hint given by the lady in the bar?
Why does Captain Marvel assume the people on this planet know this?
What wound would be of little consequence to a biped but terrible for a quadruped?
Virginia employer terminated employee and wants signing bonus returned
What is the magic ball of every day?
Can I pump my MTB tire to max (55 psi / 380 kPa) without the tube inside bursting?
meaning and function of 幸 in "则幸分我一杯羹"
Conservation of Mass and Energy
Can one live in the U.S. and not use a credit card?
Accountant/ lawyer will not return my call
Is it necessary to separate DC power cables and data cables?
Dropdown com clique
PTIJ: wiping amalek’s memory?
Is it possible to avoid unpacking when merging Association?
Do items de-spawn in Diablo?
Do I really need to have a scientific explanation for my premise?
How can The Temple of Elementary Evil reliably protect itself against kinetic bombardment?
How do I express some one as a black person?
NASA's RS-25 Engines shut down time
Recommendation letter by significant other if you worked with them professionally?
Reverse string, can I make it faster?
Declaring and defining template, and specialising them
Does "Until when" sound natural for native speakers?
Could you please stop shuffling the deck and play already?
Maximum and minimum of $x^2+y^2-2x$
Farthest and Nearest Point in the ellipse from a focus using DerivativesMin/Max of $f(x,y) = e^{xy}$ where $x^3+y^3=16$Showing that $ frac{a}{a^2+3}+frac{b}{b^2+3}leqfrac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalitiesHow to solve the inequality $frac {5x+1}{4x-1}geq1$If $a,b,c$ and $d$ are positive reals such that $a+b+c+d=1$.Find the minimum value of $frac{1}{a}+frac{1}{b}+frac{1}{c}+frac{1}{d}$Find the maximum value of $xyzw$ if $x+2y+4z+8w=16$Lagrange multiplier with further constraints on the variables.What is a boundary point when using Lagrange Multipliers?About using lagrange multiplier for the min value of $3x+4y$ when $x^2+y^2leq 16$ and $x,y in mathbb{R}$Finding the maximum of $p^3 + q^3 +r^3 + 4pqr$Proving Absolute Value Inequality Using Lagrange Multipliers
$begingroup$
Maximum and minimum value of $x^2+y^2-2x$ for $21x^2-6xy+29y^2+6x-58y-151=0$
What I tried is:
The Lagrange multiplier theorem.
$$Lambda (x,y,lambda)=f(x,y)+lambda g(x,y)=x^2+y^2-2x+lambda(21x^2-6xy+29y^2+6x-58y-151)$$
then solve the following system of equation
$$left{begin{matrix}
cfrac{dLambda}{dx}=0&&&&&&(1)\
cfrac{dLambda}{dy}=0&&&&&&(2)\
cfrac{dLambda}{dlambda}=0&&&&&&(3)\
end{matrix} right.$$
Can I solve it without Lagrange method? Help me please.
inequality
edited 2 days ago
Math Bob
8311
asked 2 days ago
jackyjacky
1,028715
$endgroup$
add a comment |
$begingroup$
Maximum and minimum value of $x^2+y^2-2x$ for $21x^2-6xy+29y^2+6x-58y-151=0$
What I tried is:
The Lagrange multiplier theorem.
$$Lambda (x,y,lambda)=f(x,y)+lambda g(x,y)=x^2+y^2-2x+lambda(21x^2-6xy+29y^2+6x-58y-151)$$
then solve the following system of equation
$$left{begin{matrix}
cfrac{dLambda}{dx}=0&&&&&&(1)\
cfrac{dLambda}{dy}=0&&&&&&(2)\
cfrac{dLambda}{dlambda}=0&&&&&&(3)\
end{matrix} right.$$
Can I solve it without Lagrange method? Help me please.
inequality
edited 2 days ago
Math Bob
8311
asked 2 days ago
jackyjacky
1,028715
$endgroup$
$begingroup$
Yes, you can, but you need to choose another coefficients.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
Maximum and minimum value of $x^2+y^2-2x$ for $21x^2-6xy+29y^2+6x-58y-151=0$
What I tried is:
The Lagrange multiplier theorem.
$$Lambda (x,y,lambda)=f(x,y)+lambda g(x,y)=x^2+y^2-2x+lambda(21x^2-6xy+29y^2+6x-58y-151)$$
then solve the following system of equation
$$left{begin{matrix}
cfrac{dLambda}{dx}=0&&&&&&(1)\
cfrac{dLambda}{dy}=0&&&&&&(2)\
cfrac{dLambda}{dlambda}=0&&&&&&(3)\
end{matrix} right.$$
Can I solve it without Lagrange method? Help me please.
inequality
edited 2 days ago
Math Bob
8311
asked 2 days ago
jackyjacky
1,028715
$endgroup$
Maximum and minimum value of $x^2+y^2-2x$ for $21x^2-6xy+29y^2+6x-58y-151=0$
What I tried is:
The Lagrange multiplier theorem.
$$Lambda (x,y,lambda)=f(x,y)+lambda g(x,y)=x^2+y^2-2x+lambda(21x^2-6xy+29y^2+6x-58y-151)$$
then solve the following system of equation
$$left{begin{matrix}
cfrac{dLambda}{dx}=0&&&&&&(1)\
cfrac{dLambda}{dy}=0&&&&&&(2)\
cfrac{dLambda}{dlambda}=0&&&&&&(3)\
end{matrix} right.$$
Can I solve it without Lagrange method? Help me please.
inequality
inequality
edited 2 days ago
Math Bob
8311
asked 2 days ago
jackyjacky
1,028715
edited 2 days ago
Math Bob
8311
asked 2 days ago
jackyjacky
1,028715
edited 2 days ago
Math Bob
8311
edited 2 days ago
Math Bob
8311
edited 2 days ago
Math Bob
8311
8311
asked 2 days ago
jackyjacky
1,028715
asked 2 days ago
jackyjacky
1,028715
asked 2 days ago
jackyjacky
1,028715
1,028715
$begingroup$
Yes, you can, but you need to choose another coefficients.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
Yes, you can, but you need to choose another coefficients.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
Yes, you can, but you need to choose another coefficients.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
Yes, you can, but you need to choose another coefficients.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, this can be done without Lagrange's method. Let us define, $ f(x,y) := 21x^2-6xy+29y^2+6x-58y-151=0$ to be your constraint. Further, let us change the coordinates to $X:=x-1$, $Y:=y$. Then your optimization problem is to find the minimum and maximum of $X^2 + Y^2 - 1$ or equivalently, of $X^2 + Y^2$ such that $f(X+1, Y)=0$. The key here is to note that $f(x,y)$ (and hence $f(X+1, Y)=0$) is a shifted ellipse (the constraint is of the form of a conic section, and you can prove that this is an ellipse; alternatively, see Wolfram Alpha), and we are looking for the point on this ellipse which is closest and furthest away from the origin. This can be done through various means.
In particular, you can look for points on the ellipse such that the normal through those points passes through the origin (this is equivalent to finding a circle through the origin which is tangential to the ellipse).
Or you can also parameterize your ellipse as has been done here.
answered 2 days ago
NoelNoel
827
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3140763%2fmaximum-and-minimum-of-x2y2-2x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, this can be done without Lagrange's method. Let us define, $ f(x,y) := 21x^2-6xy+29y^2+6x-58y-151=0$ to be your constraint. Further, let us change the coordinates to $X:=x-1$, $Y:=y$. Then your optimization problem is to find the minimum and maximum of $X^2 + Y^2 - 1$ or equivalently, of $X^2 + Y^2$ such that $f(X+1, Y)=0$. The key here is to note that $f(x,y)$ (and hence $f(X+1, Y)=0$) is a shifted ellipse (the constraint is of the form of a conic section, and you can prove that this is an ellipse; alternatively, see Wolfram Alpha), and we are looking for the point on this ellipse which is closest and furthest away from the origin. This can be done through various means.
In particular, you can look for points on the ellipse such that the normal through those points passes through the origin (this is equivalent to finding a circle through the origin which is tangential to the ellipse).
Or you can also parameterize your ellipse as has been done here.
answered 2 days ago
NoelNoel
827
$endgroup$
add a comment |
$begingroup$
Yes, this can be done without Lagrange's method. Let us define, $ f(x,y) := 21x^2-6xy+29y^2+6x-58y-151=0$ to be your constraint. Further, let us change the coordinates to $X:=x-1$, $Y:=y$. Then your optimization problem is to find the minimum and maximum of $X^2 + Y^2 - 1$ or equivalently, of $X^2 + Y^2$ such that $f(X+1, Y)=0$. The key here is to note that $f(x,y)$ (and hence $f(X+1, Y)=0$) is a shifted ellipse (the constraint is of the form of a conic section, and you can prove that this is an ellipse; alternatively, see Wolfram Alpha), and we are looking for the point on this ellipse which is closest and furthest away from the origin. This can be done through various means.
In particular, you can look for points on the ellipse such that the normal through those points passes through the origin (this is equivalent to finding a circle through the origin which is tangential to the ellipse).
Or you can also parameterize your ellipse as has been done here.
answered 2 days ago
NoelNoel
827
$endgroup$
add a comment |
$begingroup$
Yes, this can be done without Lagrange's method. Let us define, $ f(x,y) := 21x^2-6xy+29y^2+6x-58y-151=0$ to be your constraint. Further, let us change the coordinates to $X:=x-1$, $Y:=y$. Then your optimization problem is to find the minimum and maximum of $X^2 + Y^2 - 1$ or equivalently, of $X^2 + Y^2$ such that $f(X+1, Y)=0$. The key here is to note that $f(x,y)$ (and hence $f(X+1, Y)=0$) is a shifted ellipse (the constraint is of the form of a conic section, and you can prove that this is an ellipse; alternatively, see Wolfram Alpha), and we are looking for the point on this ellipse which is closest and furthest away from the origin. This can be done through various means.
In particular, you can look for points on the ellipse such that the normal through those points passes through the origin (this is equivalent to finding a circle through the origin which is tangential to the ellipse).
Or you can also parameterize your ellipse as has been done here.
answered 2 days ago
NoelNoel
827
$endgroup$
Yes, this can be done without Lagrange's method. Let us define, $ f(x,y) := 21x^2-6xy+29y^2+6x-58y-151=0$ to be your constraint. Further, let us change the coordinates to $X:=x-1$, $Y:=y$. Then your optimization problem is to find the minimum and maximum of $X^2 + Y^2 - 1$ or equivalently, of $X^2 + Y^2$ such that $f(X+1, Y)=0$. The key here is to note that $f(x,y)$ (and hence $f(X+1, Y)=0$) is a shifted ellipse (the constraint is of the form of a conic section, and you can prove that this is an ellipse; alternatively, see Wolfram Alpha), and we are looking for the point on this ellipse which is closest and furthest away from the origin. This can be done through various means.
In particular, you can look for points on the ellipse such that the normal through those points passes through the origin (this is equivalent to finding a circle through the origin which is tangential to the ellipse).
Or you can also parameterize your ellipse as has been done here.
answered 2 days ago
NoelNoel
827
answered 2 days ago
NoelNoel
827
answered 2 days ago
NoelNoel
827
answered 2 days ago
NoelNoel
827
827
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3140763%2fmaximum-and-minimum-of-x2y2-2x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Yes, you can, but you need to choose another coefficients.
$endgroup$
– Michael Rozenberg
2 days ago