Mixing
Find maximum of $cos(x)cos(y)cos(z)$ when $x + y + z = frac{pi}{2}$ and $x, y, z > 0$
$begingroup$
Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.
Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.
real-analysis trigonometry optimization maxima-minima jensen-inequality
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
$endgroup$
add a comment |
$begingroup$
Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.
Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.
real-analysis trigonometry optimization maxima-minima jensen-inequality
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
$endgroup$
add a comment |
$begingroup$
Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.
Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.
real-analysis trigonometry optimization maxima-minima jensen-inequality
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
$endgroup$
Find maximum value of P = $cos(x)cos(y)cos(z)$, given that $x + y + z = frac{pi}{2}$ and $x, y, z > 0$.
Effort 1. I drew a quarter-circle, divided the square angle into three parts, and attempted to derive the expression, but it went to nowhere.
real-analysis trigonometry optimization maxima-minima jensen-inequality
real-analysis trigonometry optimization maxima-minima jensen-inequality
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
edited Feb 2 at 12:39
Michael Rozenberg
110k1896201
110k1896201
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
asked Feb 2 at 5:49
KiWiJuiceKiWiJuice
186
186
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $f(x)=lncos{x}.$
Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
$$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
A geometric approach:
Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.
Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.
And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.
Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
$endgroup$
$begingroup$
Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
$endgroup$
– KiWiJuice
Mar 14 at 7:44
$begingroup$
While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
$endgroup$
– jmerry
Mar 14 at 7:48
add a comment |
Your Answer
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%2f3097058%2ffind-maximum-of-cosx-cosy-cosz-when-x-y-z-frac-pi2-and-x%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 $f(x)=lncos{x}.$
Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
$$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Let $f(x)=lncos{x}.$
Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
$$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Let $f(x)=lncos{x}.$
Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
$$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
Let $f(x)=lncos{x}.$
Thus, $$f''(x)=-left(tan{x}right)'=-frac{1}{cos^2x}<0.$$ Thus, $f$ is a concave function and since $e^x$ increases, by Jensen we obtain:
$$cos{x}cos{y}cos{z}=e^{sumlimits_{cyc}lncos{x}}leq e^{3lncosfrac{x+y+z}{3}}=frac{3sqrt3}{8}.$$
The equality occurs for $x=y=z=frac{pi}{6},$ which says that we got a maximal value.
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
edited Feb 2 at 5:57
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 2 at 5:52
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
$begingroup$
A geometric approach:
Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.
Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.
And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.
Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
$endgroup$
$begingroup$
Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
$endgroup$
– KiWiJuice
Mar 14 at 7:44
$begingroup$
While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
$endgroup$
– jmerry
Mar 14 at 7:48
add a comment |
$begingroup$
A geometric approach:
Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.
Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.
And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.
Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
$endgroup$
$begingroup$
Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
$endgroup$
– KiWiJuice
Mar 14 at 7:44
$begingroup$
While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
$endgroup$
– jmerry
Mar 14 at 7:48
add a comment |
$begingroup$
A geometric approach:
Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.
Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.
And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.
Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
$endgroup$
A geometric approach:
Consider an acute triangle $ABC$ with angles labeled by their vertices, and set $x=frac{pi}{2}-A$, $y=frac{pi}{2}-B$, $z=frac{pi}{2}-C$. Since $A+B+C=pi$, $x+y+z=frac{pi}{2}$. The acute triangle restriction ensures that $x,y,z>0$, and the restrictions are the same. In terms of $A,B,C$, the quantity we seek to maximize is $sin Asin Bsin C$.
Now, we get clever. The area of triangle $ABC$ is equal to $2R^2sin Asin Bsin C$, where $R$ is the circumradius of the triangle. Why? Because $2Rsin A=a$ and $2Rsin B=b$, we can reduce to the more familiar area formula $frac12absin C$. Maximizing $sin Asin Bsin C$ is now equivalent to maximizing the area of an acute triangle inscribed in a fixed circle $Gamma$.
And to do that, let's loosen things a bit, and instead maximize the area of any triangle inscribed in $Gamma$. If we fix two points $A$ and $B$ on the circle, what should $C$ be to maximize the area? It should be as far away from the line $AB$ as possible, namely one of the endpoints of the diameter of $Gamma$ that is the perpendicular bisector of $AB$. Comparing the two endpoints, the farther one is the one with the acute angle (or, if they're both right angles, both are equally far), so the area maximum with $A$ and $B$ fixed has $a=b$ and angle $C$ acute. Repeat for the other two pairs of vertices, and the maximum must come when $a=b=c$ and the triangle is equilateral. That's $60^circ$ angles $A,B,C$, for $x=y=z=frac{pi}{6}$ and $cos xcos ycos z = sin Asin Bsin C=left(frac{sqrt{3}}{2}right)^3=frac{3sqrt{3}}{8}$.
Now, that's a little unsatisfying logically - we proved that the maximum, if it exists, must be at an equilateral triangle. To patch this, we prove that there is a maximum. The area is a continuous function from $Gamma^3$ (the positions of the three points) to $mathbb{R}$. That's a compact domain, so the extreme value theorem applies and there is a maximum. Done.
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
answered Feb 2 at 6:34
jmerryjmerry
17.1k11633
17.1k11633
$begingroup$
Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
$endgroup$
– KiWiJuice
Mar 14 at 7:44
$begingroup$
While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
$endgroup$
– jmerry
Mar 14 at 7:48
add a comment |
$begingroup$
Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
$endgroup$
– KiWiJuice
Mar 14 at 7:44
$begingroup$
While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
$endgroup$
– jmerry
Mar 14 at 7:48
$begingroup$
Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
$endgroup$
– KiWiJuice
Mar 14 at 7:44
$begingroup$
Thanks for the good idea. But can we generalize this approach to n terms instead of just only 3 terms?
$endgroup$
– KiWiJuice
Mar 14 at 7:44
$begingroup$
While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
$endgroup$
– jmerry
Mar 14 at 7:48
$begingroup$
While we could do something based on polygons with more sides, it would look very different - not a simple product of cosines. This argument depends on the area formula $K=2R^2sin Asin Bsin C$, which is specific to triangles.
$endgroup$
– jmerry
Mar 14 at 7:48
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%2f3097058%2ffind-maximum-of-cosx-cosy-cosz-when-x-y-z-frac-pi2-and-x%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
