Calculus Optimization Problem: Wire Triangle and Circle
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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 at 4:33
Jasmine A
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favorite
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 at 4:33
Jasmine A
1
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 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 at 4:33
Jasmine A
1
asked Nov 21 at 4:33
Jasmine A
1
asked Nov 21 at 4:33
Jasmine A
1
asked Nov 21 at 4:33
Jasmine A
1
asked Nov 21 at 4:33
Jasmine A
1
1
add a comment |
add a comment |
2 Answers
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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 at 4:47
Offlaw
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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 at 4:55
Yadati Kiran
1,428518
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2 Answers
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2 Answers
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up vote
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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 at 4:47
Offlaw
2649
add a comment |
up vote
0
down vote
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 at 4:47
Offlaw
2649
add a comment |
up vote
0
down vote
up vote
0
down vote
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 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 at 4:47
Offlaw
2649
answered Nov 21 at 4:47
Offlaw
2649
answered Nov 21 at 4:47
Offlaw
2649
answered Nov 21 at 4:47
Offlaw
2649
2649
add a comment |
add a comment |
up vote
0
down vote
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 at 4:55
Yadati Kiran
1,428518
add a comment |
up vote
0
down vote
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 at 4:55
Yadati Kiran
1,428518
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 4:55
Yadati Kiran
1,428518
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 at 4:55
Yadati Kiran
1,428518
edited Nov 21 at 5:01
answered Nov 21 at 4:55
Yadati Kiran
1,428518
answered Nov 21 at 4:55
Yadati Kiran
1,428518
answered Nov 21 at 4:55
Yadati Kiran
1,428518
1,428518
add a comment |
add a comment |
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