Mixing
Calculus Optimization Problem: Wire Triangle and Circle
A wire 5 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each.
I understand most of this but I'm having trouble finding the right terms for the triangle's area and optimization. Any help would be appreciated.
calculus optimization triangle circle area
asked Nov 21 '18 at 4:33
Jasmine A
1
add a comment |
A wire 5 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each.
I understand most of this but I'm having trouble finding the right terms for the triangle's area and optimization. Any help would be appreciated.
calculus optimization triangle circle area
asked Nov 21 '18 at 4:33
Jasmine A
1
add a comment |
A wire 5 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each.
I understand most of this but I'm having trouble finding the right terms for the triangle's area and optimization. Any help would be appreciated.
calculus optimization triangle circle area
asked Nov 21 '18 at 4:33
Jasmine A
1
A wire 5 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each.
I understand most of this but I'm having trouble finding the right terms for the triangle's area and optimization. Any help would be appreciated.
calculus optimization triangle circle area
calculus optimization triangle circle area
asked Nov 21 '18 at 4:33
Jasmine A
1
asked Nov 21 '18 at 4:33
Jasmine A
1
asked Nov 21 '18 at 4:33
Jasmine A
1
asked Nov 21 '18 at 4:33
Jasmine A
1
asked Nov 21 '18 at 4:33
Jasmine A
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Consider $x=$ the length of the part used to make equilateral triangle.
Then each side of the triangle $= x/3$ and area of the triangle $= frac{surd{3}x^2}{36}$.
Area of the circle $=pi(frac{5-x}{2pi})^2$
Let $f(x)=frac{surd{3}x^2}{36}+pi(frac{5-x}{2pi})^2$
Let $f'(x)=frac{surd{3}x}{18}-frac{5-x}{2pi}=0$, then find $x$.
Also $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$
answered Nov 21 '18 at 4:47
Offlaw
2649
add a comment |
Let $x$ an $5-x$ be the length of wires and bent into an equilateral triangle and a circle respectively. Area of equilateral triangle given side length $dfrac{x}{3}$ is $dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)$ and that of circle given circumference $5-x$ is $dfrac{(5-x)^2}{4pi}$. Total area $f(x)=dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)+dfrac{(5-x)^2}{4pi}$. To minimize we differentiate, find the critical points and determine the local minima. $f'(x)=dfrac{sqrt{3}}{18}x+dfrac{(x-5)}{2pi}=0$ we get
$x=dfrac{45}{9+pisqrt{3}}$. By second derivative test we see minima occurs at $x=dfrac{45}{9+pisqrt{3}}approx3.11604$
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
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%2f3007265%2fcalculus-optimization-problem-wire-triangle-and-circle%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
Consider $x=$ the length of the part used to make equilateral triangle.
Then each side of the triangle $= x/3$ and area of the triangle $= frac{surd{3}x^2}{36}$.
Area of the circle $=pi(frac{5-x}{2pi})^2$
Let $f(x)=frac{surd{3}x^2}{36}+pi(frac{5-x}{2pi})^2$
Let $f'(x)=frac{surd{3}x}{18}-frac{5-x}{2pi}=0$, then find $x$.
Also $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$
answered Nov 21 '18 at 4:47
Offlaw
2649
add a comment |
Consider $x=$ the length of the part used to make equilateral triangle.
Then each side of the triangle $= x/3$ and area of the triangle $= frac{surd{3}x^2}{36}$.
Area of the circle $=pi(frac{5-x}{2pi})^2$
Let $f(x)=frac{surd{3}x^2}{36}+pi(frac{5-x}{2pi})^2$
Let $f'(x)=frac{surd{3}x}{18}-frac{5-x}{2pi}=0$, then find $x$.
Also $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$
answered Nov 21 '18 at 4:47
Offlaw
2649
add a comment |
Consider $x=$ the length of the part used to make equilateral triangle.
Then each side of the triangle $= x/3$ and area of the triangle $= frac{surd{3}x^2}{36}$.
Area of the circle $=pi(frac{5-x}{2pi})^2$
Let $f(x)=frac{surd{3}x^2}{36}+pi(frac{5-x}{2pi})^2$
Let $f'(x)=frac{surd{3}x}{18}-frac{5-x}{2pi}=0$, then find $x$.
Also $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$
answered Nov 21 '18 at 4:47
Offlaw
2649
Consider $x=$ the length of the part used to make equilateral triangle.
Then each side of the triangle $= x/3$ and area of the triangle $= frac{surd{3}x^2}{36}$.
Area of the circle $=pi(frac{5-x}{2pi})^2$
Let $f(x)=frac{surd{3}x^2}{36}+pi(frac{5-x}{2pi})^2$
Let $f'(x)=frac{surd{3}x}{18}-frac{5-x}{2pi}=0$, then find $x$.
Also $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$
answered Nov 21 '18 at 4:47
Offlaw
2649
answered Nov 21 '18 at 4:47
Offlaw
2649
answered Nov 21 '18 at 4:47
Offlaw
2649
answered Nov 21 '18 at 4:47
Offlaw
2649
2649
add a comment |
add a comment |
Let $x$ an $5-x$ be the length of wires and bent into an equilateral triangle and a circle respectively. Area of equilateral triangle given side length $dfrac{x}{3}$ is $dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)$ and that of circle given circumference $5-x$ is $dfrac{(5-x)^2}{4pi}$. Total area $f(x)=dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)+dfrac{(5-x)^2}{4pi}$. To minimize we differentiate, find the critical points and determine the local minima. $f'(x)=dfrac{sqrt{3}}{18}x+dfrac{(x-5)}{2pi}=0$ we get
$x=dfrac{45}{9+pisqrt{3}}$. By second derivative test we see minima occurs at $x=dfrac{45}{9+pisqrt{3}}approx3.11604$
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
add a comment |
Let $x$ an $5-x$ be the length of wires and bent into an equilateral triangle and a circle respectively. Area of equilateral triangle given side length $dfrac{x}{3}$ is $dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)$ and that of circle given circumference $5-x$ is $dfrac{(5-x)^2}{4pi}$. Total area $f(x)=dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)+dfrac{(5-x)^2}{4pi}$. To minimize we differentiate, find the critical points and determine the local minima. $f'(x)=dfrac{sqrt{3}}{18}x+dfrac{(x-5)}{2pi}=0$ we get
$x=dfrac{45}{9+pisqrt{3}}$. By second derivative test we see minima occurs at $x=dfrac{45}{9+pisqrt{3}}approx3.11604$
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
add a comment |
Let $x$ an $5-x$ be the length of wires and bent into an equilateral triangle and a circle respectively. Area of equilateral triangle given side length $dfrac{x}{3}$ is $dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)$ and that of circle given circumference $5-x$ is $dfrac{(5-x)^2}{4pi}$. Total area $f(x)=dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)+dfrac{(5-x)^2}{4pi}$. To minimize we differentiate, find the critical points and determine the local minima. $f'(x)=dfrac{sqrt{3}}{18}x+dfrac{(x-5)}{2pi}=0$ we get
$x=dfrac{45}{9+pisqrt{3}}$. By second derivative test we see minima occurs at $x=dfrac{45}{9+pisqrt{3}}approx3.11604$
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
Let $x$ an $5-x$ be the length of wires and bent into an equilateral triangle and a circle respectively. Area of equilateral triangle given side length $dfrac{x}{3}$ is $dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)$ and that of circle given circumference $5-x$ is $dfrac{(5-x)^2}{4pi}$. Total area $f(x)=dfrac{sqrt{3}}{4}left(dfrac{x^2}{9}right)+dfrac{(5-x)^2}{4pi}$. To minimize we differentiate, find the critical points and determine the local minima. $f'(x)=dfrac{sqrt{3}}{18}x+dfrac{(x-5)}{2pi}=0$ we get
$x=dfrac{45}{9+pisqrt{3}}$. By second derivative test we see minima occurs at $x=dfrac{45}{9+pisqrt{3}}approx3.11604$
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
edited Nov 21 '18 at 5:01
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
answered Nov 21 '18 at 4:55
Yadati Kiran
1,693619
1,693619
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3007265%2fcalculus-optimization-problem-wire-triangle-and-circle%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
