Mixing
General maximization/minimization technique without calculus?
$begingroup$
I know that you can use derivatives to find the minimum/maximum of a function. However, sometimes it is very difficult or too long. For example,
In the problem above, there are 3 variables, and on the AMC 10 you don't exactly have the time for fancy calculus. Is there another general technique for cases like this? Thanks!
calculus optimization contest-math
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
asked Jan 31 at 19:10
NikosNikos
32
$endgroup$
add a comment |
$begingroup$
I know that you can use derivatives to find the minimum/maximum of a function. However, sometimes it is very difficult or too long. For example,
In the problem above, there are 3 variables, and on the AMC 10 you don't exactly have the time for fancy calculus. Is there another general technique for cases like this? Thanks!
calculus optimization contest-math
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
asked Jan 31 at 19:10
NikosNikos
32
$endgroup$
1
$begingroup$
You would not be able to solve this problem by calculus, because it is an integer problem.
$endgroup$
– Gerhard S.
Jan 31 at 19:18
add a comment |
$begingroup$
I know that you can use derivatives to find the minimum/maximum of a function. However, sometimes it is very difficult or too long. For example,
In the problem above, there are 3 variables, and on the AMC 10 you don't exactly have the time for fancy calculus. Is there another general technique for cases like this? Thanks!
calculus optimization contest-math
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
asked Jan 31 at 19:10
NikosNikos
32
$endgroup$
I know that you can use derivatives to find the minimum/maximum of a function. However, sometimes it is very difficult or too long. For example,
In the problem above, there are 3 variables, and on the AMC 10 you don't exactly have the time for fancy calculus. Is there another general technique for cases like this? Thanks!
calculus optimization contest-math
calculus optimization contest-math
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
asked Jan 31 at 19:10
NikosNikos
32
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
asked Jan 31 at 19:10
NikosNikos
32
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
edited Feb 1 at 4:45
Rodrigo de Azevedo
13.1k41962
13.1k41962
asked Jan 31 at 19:10
NikosNikos
32
asked Jan 31 at 19:10
NikosNikos
32
asked Jan 31 at 19:10
NikosNikos
32
32
1
$begingroup$
You would not be able to solve this problem by calculus, because it is an integer problem.
$endgroup$
– Gerhard S.
Jan 31 at 19:18
add a comment |
1
$begingroup$
You would not be able to solve this problem by calculus, because it is an integer problem.
$endgroup$
– Gerhard S.
Jan 31 at 19:18
1
1
$begingroup$
You would not be able to solve this problem by calculus, because it is an integer problem.
$endgroup$
– Gerhard S.
Jan 31 at 19:18
$begingroup$
You would not be able to solve this problem by calculus, because it is an integer problem.
$endgroup$
– Gerhard S.
Jan 31 at 19:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $S = AMC+AM+AC+MC$, and let $x=A+1$, $y=M+1$, $z=C+1$. We notice that
$$xyz = S+A+M+C+1 = S+10+1=S+11$$
so we can maximize $S$ by maximizing $xyz$ subject to the constraint $x+y+z =13$, since $A+1+M+1+C+1=13$.
For the moment let's suppose $x,y,z$ can be non-integer. Then by the AM-GM inequality, we know $xyz$ is maximized when $x=y=z$. Since $x+y+z=13$, this yields $x=y=z=13/3$, with $xyz=(13/3)^3 approx 81.4$, corresponding to $S =xyz-11approx 70.4$. So with integer constraints, we know we can't do better than $S=70$; that eliminates choices (D) and (E).
If we look for integer values near $13/3$ with sum $13$, it's natural to try $x=y=4$, $z=5$, which yields $xyz=80$, corresponding to $S=80-11=69$. So the answer is (C).
answered Feb 1 at 16:53
awkwardawkward
6,89011026
$endgroup$
add a comment |
$begingroup$
For this specific problem, I just thought, that amc + am + mc + ac thing looks a lot like multiplication, I bet it behaves like that. And when splitting an interval to make a product (choose a + b = 1 to maximize ab) the maximal split is half-and-half, so I figured for splitting it into 3 parts to make a product the maximal split would be around third-third-third. So choose 3-3-4. And lo and behold, it gives one of the answers.
With multiple choice, you can kinda just guess.
answered Feb 1 at 6:07
user640459user640459
111
$endgroup$
$begingroup$
Welcome to math.stackexchange. So far, your answer contains some heuristic arguments for why 3-3-4 might be the best solution for obtaining a maximum value. I think your answer should also contain a rigorous proof why 3-3-4 is the best solution.
$endgroup$
– supinf
Feb 1 at 6:52
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%2f3095329%2fgeneral-maximization-minimization-technique-without-calculus%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
$begingroup$
Let $S = AMC+AM+AC+MC$, and let $x=A+1$, $y=M+1$, $z=C+1$. We notice that
$$xyz = S+A+M+C+1 = S+10+1=S+11$$
so we can maximize $S$ by maximizing $xyz$ subject to the constraint $x+y+z =13$, since $A+1+M+1+C+1=13$.
For the moment let's suppose $x,y,z$ can be non-integer. Then by the AM-GM inequality, we know $xyz$ is maximized when $x=y=z$. Since $x+y+z=13$, this yields $x=y=z=13/3$, with $xyz=(13/3)^3 approx 81.4$, corresponding to $S =xyz-11approx 70.4$. So with integer constraints, we know we can't do better than $S=70$; that eliminates choices (D) and (E).
If we look for integer values near $13/3$ with sum $13$, it's natural to try $x=y=4$, $z=5$, which yields $xyz=80$, corresponding to $S=80-11=69$. So the answer is (C).
answered Feb 1 at 16:53
awkwardawkward
6,89011026
$endgroup$
add a comment |
$begingroup$
Let $S = AMC+AM+AC+MC$, and let $x=A+1$, $y=M+1$, $z=C+1$. We notice that
$$xyz = S+A+M+C+1 = S+10+1=S+11$$
so we can maximize $S$ by maximizing $xyz$ subject to the constraint $x+y+z =13$, since $A+1+M+1+C+1=13$.
For the moment let's suppose $x,y,z$ can be non-integer. Then by the AM-GM inequality, we know $xyz$ is maximized when $x=y=z$. Since $x+y+z=13$, this yields $x=y=z=13/3$, with $xyz=(13/3)^3 approx 81.4$, corresponding to $S =xyz-11approx 70.4$. So with integer constraints, we know we can't do better than $S=70$; that eliminates choices (D) and (E).
If we look for integer values near $13/3$ with sum $13$, it's natural to try $x=y=4$, $z=5$, which yields $xyz=80$, corresponding to $S=80-11=69$. So the answer is (C).
answered Feb 1 at 16:53
awkwardawkward
6,89011026
$endgroup$
add a comment |
$begingroup$
Let $S = AMC+AM+AC+MC$, and let $x=A+1$, $y=M+1$, $z=C+1$. We notice that
$$xyz = S+A+M+C+1 = S+10+1=S+11$$
so we can maximize $S$ by maximizing $xyz$ subject to the constraint $x+y+z =13$, since $A+1+M+1+C+1=13$.
For the moment let's suppose $x,y,z$ can be non-integer. Then by the AM-GM inequality, we know $xyz$ is maximized when $x=y=z$. Since $x+y+z=13$, this yields $x=y=z=13/3$, with $xyz=(13/3)^3 approx 81.4$, corresponding to $S =xyz-11approx 70.4$. So with integer constraints, we know we can't do better than $S=70$; that eliminates choices (D) and (E).
If we look for integer values near $13/3$ with sum $13$, it's natural to try $x=y=4$, $z=5$, which yields $xyz=80$, corresponding to $S=80-11=69$. So the answer is (C).
answered Feb 1 at 16:53
awkwardawkward
6,89011026
$endgroup$
Let $S = AMC+AM+AC+MC$, and let $x=A+1$, $y=M+1$, $z=C+1$. We notice that
$$xyz = S+A+M+C+1 = S+10+1=S+11$$
so we can maximize $S$ by maximizing $xyz$ subject to the constraint $x+y+z =13$, since $A+1+M+1+C+1=13$.
For the moment let's suppose $x,y,z$ can be non-integer. Then by the AM-GM inequality, we know $xyz$ is maximized when $x=y=z$. Since $x+y+z=13$, this yields $x=y=z=13/3$, with $xyz=(13/3)^3 approx 81.4$, corresponding to $S =xyz-11approx 70.4$. So with integer constraints, we know we can't do better than $S=70$; that eliminates choices (D) and (E).
If we look for integer values near $13/3$ with sum $13$, it's natural to try $x=y=4$, $z=5$, which yields $xyz=80$, corresponding to $S=80-11=69$. So the answer is (C).
answered Feb 1 at 16:53
awkwardawkward
6,89011026
edited Feb 2 at 13:11
answered Feb 1 at 16:53
awkwardawkward
6,89011026
answered Feb 1 at 16:53
awkwardawkward
6,89011026
answered Feb 1 at 16:53
awkwardawkward
6,89011026
6,89011026
add a comment |
add a comment |
$begingroup$
For this specific problem, I just thought, that amc + am + mc + ac thing looks a lot like multiplication, I bet it behaves like that. And when splitting an interval to make a product (choose a + b = 1 to maximize ab) the maximal split is half-and-half, so I figured for splitting it into 3 parts to make a product the maximal split would be around third-third-third. So choose 3-3-4. And lo and behold, it gives one of the answers.
With multiple choice, you can kinda just guess.
answered Feb 1 at 6:07
user640459user640459
111
$endgroup$
$begingroup$
Welcome to math.stackexchange. So far, your answer contains some heuristic arguments for why 3-3-4 might be the best solution for obtaining a maximum value. I think your answer should also contain a rigorous proof why 3-3-4 is the best solution.
$endgroup$
– supinf
Feb 1 at 6:52
add a comment |
$begingroup$
For this specific problem, I just thought, that amc + am + mc + ac thing looks a lot like multiplication, I bet it behaves like that. And when splitting an interval to make a product (choose a + b = 1 to maximize ab) the maximal split is half-and-half, so I figured for splitting it into 3 parts to make a product the maximal split would be around third-third-third. So choose 3-3-4. And lo and behold, it gives one of the answers.
With multiple choice, you can kinda just guess.
answered Feb 1 at 6:07
user640459user640459
111
$endgroup$
$begingroup$
Welcome to math.stackexchange. So far, your answer contains some heuristic arguments for why 3-3-4 might be the best solution for obtaining a maximum value. I think your answer should also contain a rigorous proof why 3-3-4 is the best solution.
$endgroup$
– supinf
Feb 1 at 6:52
add a comment |
$begingroup$
For this specific problem, I just thought, that amc + am + mc + ac thing looks a lot like multiplication, I bet it behaves like that. And when splitting an interval to make a product (choose a + b = 1 to maximize ab) the maximal split is half-and-half, so I figured for splitting it into 3 parts to make a product the maximal split would be around third-third-third. So choose 3-3-4. And lo and behold, it gives one of the answers.
With multiple choice, you can kinda just guess.
answered Feb 1 at 6:07
user640459user640459
111
$endgroup$
For this specific problem, I just thought, that amc + am + mc + ac thing looks a lot like multiplication, I bet it behaves like that. And when splitting an interval to make a product (choose a + b = 1 to maximize ab) the maximal split is half-and-half, so I figured for splitting it into 3 parts to make a product the maximal split would be around third-third-third. So choose 3-3-4. And lo and behold, it gives one of the answers.
With multiple choice, you can kinda just guess.
answered Feb 1 at 6:07
user640459user640459
111
answered Feb 1 at 6:07
user640459user640459
111
answered Feb 1 at 6:07
user640459user640459
111
answered Feb 1 at 6:07
user640459user640459
111
111
$begingroup$
Welcome to math.stackexchange. So far, your answer contains some heuristic arguments for why 3-3-4 might be the best solution for obtaining a maximum value. I think your answer should also contain a rigorous proof why 3-3-4 is the best solution.
$endgroup$
– supinf
Feb 1 at 6:52
add a comment |
$begingroup$
Welcome to math.stackexchange. So far, your answer contains some heuristic arguments for why 3-3-4 might be the best solution for obtaining a maximum value. I think your answer should also contain a rigorous proof why 3-3-4 is the best solution.
$endgroup$
– supinf
Feb 1 at 6:52
$begingroup$
Welcome to math.stackexchange. So far, your answer contains some heuristic arguments for why 3-3-4 might be the best solution for obtaining a maximum value. I think your answer should also contain a rigorous proof why 3-3-4 is the best solution.
$endgroup$
– supinf
Feb 1 at 6:52
$begingroup$
Welcome to math.stackexchange. So far, your answer contains some heuristic arguments for why 3-3-4 might be the best solution for obtaining a maximum value. I think your answer should also contain a rigorous proof why 3-3-4 is the best solution.
$endgroup$
– supinf
Feb 1 at 6:52
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%2f3095329%2fgeneral-maximization-minimization-technique-without-calculus%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
1
$begingroup$
You would not be able to solve this problem by calculus, because it is an integer problem.
$endgroup$
– Gerhard S.
Jan 31 at 19:18