Maximum / minimum values of $z = x + |x - y|$ with constraints
$begingroup$
Given a function $z$, with constraints $|x|le 1$, $|y|leq2$, find its minimum and maximum $$ z = x + |x-y|$$
...
$$frac{partial f(x, y)}{partial x} = frac{x-y}{|x-y|} + 1$$
$$frac{partial f(x, y)}{partial y} = frac{y-x}{|x-y|}$$
Thus, it yields a system of equations, and by solving it we're to find function's extrema points:
begin{cases} frac{x-y}{|x-y|} + 1 = 0 \ frac{y-x}{|x-y|} = 0 end{cases}
I was unable to derive any solutions by brute force, and it is not quite clear to me if it is possible to do that with analytic methods. How to proceed?
multivariable-calculus maxima-minima
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
$endgroup$
add a comment |
$begingroup$
Given a function $z$, with constraints $|x|le 1$, $|y|leq2$, find its minimum and maximum $$ z = x + |x-y|$$
...
$$frac{partial f(x, y)}{partial x} = frac{x-y}{|x-y|} + 1$$
$$frac{partial f(x, y)}{partial y} = frac{y-x}{|x-y|}$$
Thus, it yields a system of equations, and by solving it we're to find function's extrema points:
begin{cases} frac{x-y}{|x-y|} + 1 = 0 \ frac{y-x}{|x-y|} = 0 end{cases}
I was unable to derive any solutions by brute force, and it is not quite clear to me if it is possible to do that with analytic methods. How to proceed?
multivariable-calculus maxima-minima
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
$endgroup$
add a comment |
$begingroup$
Given a function $z$, with constraints $|x|le 1$, $|y|leq2$, find its minimum and maximum $$ z = x + |x-y|$$
...
$$frac{partial f(x, y)}{partial x} = frac{x-y}{|x-y|} + 1$$
$$frac{partial f(x, y)}{partial y} = frac{y-x}{|x-y|}$$
Thus, it yields a system of equations, and by solving it we're to find function's extrema points:
begin{cases} frac{x-y}{|x-y|} + 1 = 0 \ frac{y-x}{|x-y|} = 0 end{cases}
I was unable to derive any solutions by brute force, and it is not quite clear to me if it is possible to do that with analytic methods. How to proceed?
multivariable-calculus maxima-minima
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
$endgroup$
Given a function $z$, with constraints $|x|le 1$, $|y|leq2$, find its minimum and maximum $$ z = x + |x-y|$$
...
$$frac{partial f(x, y)}{partial x} = frac{x-y}{|x-y|} + 1$$
$$frac{partial f(x, y)}{partial y} = frac{y-x}{|x-y|}$$
Thus, it yields a system of equations, and by solving it we're to find function's extrema points:
begin{cases} frac{x-y}{|x-y|} + 1 = 0 \ frac{y-x}{|x-y|} = 0 end{cases}
I was unable to derive any solutions by brute force, and it is not quite clear to me if it is possible to do that with analytic methods. How to proceed?
multivariable-calculus maxima-minima
multivariable-calculus maxima-minima
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
asked Dec 15 '18 at 14:35
H. BaoH. Bao
474
474
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using calculus here is unnecessary and, given the use of absolute values, arguably inappropriate.
What you'd like to do for the maximum is simultaneously make $x$ as large as possible and $x$ and $y$ as different as possible. The constraints $|x|le1, |y|le2$ allow you to this by choosing $x=1$ and $y=-2$, for a maximum value $z=1+|1-(-2)|=4$.
For the minimum, you want to simultaneously make $x$ as small as possible and $x$ and $y$ as close as possible. This happens with $x=y=-1$, for a minimum value of $z=-1+|-1-(-1)|=-1$.
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
$endgroup$
add a comment |
$begingroup$
$$begin{align}
z&=x+|x-y|\
&geqslant x+0\
&geqslant-1
end{align}$$
And
$$begin{align}
z&=x+|x-y|\
&=x+|x+(-y)|\
&leqslant x+|x|+|-y|\
&=x+|x|+|y|\
&leqslant |x|+|x|+|y|\
&= 2|x|+|y|\
&leqslant 2+2\
&=4
end{align}$$
Can you find $(x,y)$ pairs wich for $z=-1$ and $z=4$?
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
$endgroup$
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%2f3041555%2fmaximum-minimum-values-of-z-x-x-y-with-constraints%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using calculus here is unnecessary and, given the use of absolute values, arguably inappropriate.
What you'd like to do for the maximum is simultaneously make $x$ as large as possible and $x$ and $y$ as different as possible. The constraints $|x|le1, |y|le2$ allow you to this by choosing $x=1$ and $y=-2$, for a maximum value $z=1+|1-(-2)|=4$.
For the minimum, you want to simultaneously make $x$ as small as possible and $x$ and $y$ as close as possible. This happens with $x=y=-1$, for a minimum value of $z=-1+|-1-(-1)|=-1$.
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
$endgroup$
add a comment |
$begingroup$
Using calculus here is unnecessary and, given the use of absolute values, arguably inappropriate.
What you'd like to do for the maximum is simultaneously make $x$ as large as possible and $x$ and $y$ as different as possible. The constraints $|x|le1, |y|le2$ allow you to this by choosing $x=1$ and $y=-2$, for a maximum value $z=1+|1-(-2)|=4$.
For the minimum, you want to simultaneously make $x$ as small as possible and $x$ and $y$ as close as possible. This happens with $x=y=-1$, for a minimum value of $z=-1+|-1-(-1)|=-1$.
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
$endgroup$
add a comment |
$begingroup$
Using calculus here is unnecessary and, given the use of absolute values, arguably inappropriate.
What you'd like to do for the maximum is simultaneously make $x$ as large as possible and $x$ and $y$ as different as possible. The constraints $|x|le1, |y|le2$ allow you to this by choosing $x=1$ and $y=-2$, for a maximum value $z=1+|1-(-2)|=4$.
For the minimum, you want to simultaneously make $x$ as small as possible and $x$ and $y$ as close as possible. This happens with $x=y=-1$, for a minimum value of $z=-1+|-1-(-1)|=-1$.
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
$endgroup$
Using calculus here is unnecessary and, given the use of absolute values, arguably inappropriate.
What you'd like to do for the maximum is simultaneously make $x$ as large as possible and $x$ and $y$ as different as possible. The constraints $|x|le1, |y|le2$ allow you to this by choosing $x=1$ and $y=-2$, for a maximum value $z=1+|1-(-2)|=4$.
For the minimum, you want to simultaneously make $x$ as small as possible and $x$ and $y$ as close as possible. This happens with $x=y=-1$, for a minimum value of $z=-1+|-1-(-1)|=-1$.
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
answered Dec 15 '18 at 15:13
Barry CipraBarry Cipra
59.9k654126
59.9k654126
add a comment |
add a comment |
$begingroup$
$$begin{align}
z&=x+|x-y|\
&geqslant x+0\
&geqslant-1
end{align}$$
And
$$begin{align}
z&=x+|x-y|\
&=x+|x+(-y)|\
&leqslant x+|x|+|-y|\
&=x+|x|+|y|\
&leqslant |x|+|x|+|y|\
&= 2|x|+|y|\
&leqslant 2+2\
&=4
end{align}$$
Can you find $(x,y)$ pairs wich for $z=-1$ and $z=4$?
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
$endgroup$
add a comment |
$begingroup$
$$begin{align}
z&=x+|x-y|\
&geqslant x+0\
&geqslant-1
end{align}$$
And
$$begin{align}
z&=x+|x-y|\
&=x+|x+(-y)|\
&leqslant x+|x|+|-y|\
&=x+|x|+|y|\
&leqslant |x|+|x|+|y|\
&= 2|x|+|y|\
&leqslant 2+2\
&=4
end{align}$$
Can you find $(x,y)$ pairs wich for $z=-1$ and $z=4$?
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
$endgroup$
add a comment |
$begingroup$
$$begin{align}
z&=x+|x-y|\
&geqslant x+0\
&geqslant-1
end{align}$$
And
$$begin{align}
z&=x+|x-y|\
&=x+|x+(-y)|\
&leqslant x+|x|+|-y|\
&=x+|x|+|y|\
&leqslant |x|+|x|+|y|\
&= 2|x|+|y|\
&leqslant 2+2\
&=4
end{align}$$
Can you find $(x,y)$ pairs wich for $z=-1$ and $z=4$?
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
$endgroup$
$$begin{align}
z&=x+|x-y|\
&geqslant x+0\
&geqslant-1
end{align}$$
And
$$begin{align}
z&=x+|x-y|\
&=x+|x+(-y)|\
&leqslant x+|x|+|-y|\
&=x+|x|+|y|\
&leqslant |x|+|x|+|y|\
&= 2|x|+|y|\
&leqslant 2+2\
&=4
end{align}$$
Can you find $(x,y)$ pairs wich for $z=-1$ and $z=4$?
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
answered Dec 15 '18 at 14:57
BotondBotond
5,9732832
5,9732832
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%2f3041555%2fmaximum-minimum-values-of-z-x-x-y-with-constraints%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
