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







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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







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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







      share|cite|improve this question













      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






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      asked Nov 21 '18 at 4:33









      Jasmine A

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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 $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$






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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 $x=dfrac{45}{9+pisqrt{3}}approx3.11604$






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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 $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$






              share|cite|improve this answer


























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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 $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$






                share|cite|improve this answer
























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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 $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$






                  share|cite|improve this answer












                  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 $f''(x)=frac{surd{3}}{18}+frac{1}{2pi}>0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 21 '18 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 $x=dfrac{45}{9+pisqrt{3}}approx3.11604$






                      share|cite|improve this answer




























                        0














                        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 $x=dfrac{45}{9+pisqrt{3}}approx3.11604$






                        share|cite|improve this answer


























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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 $x=dfrac{45}{9+pisqrt{3}}approx3.11604$






                          share|cite|improve this answer














                          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 $x=dfrac{45}{9+pisqrt{3}}approx3.11604$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 21 '18 at 5:01

























                          answered Nov 21 '18 at 4:55









                          Yadati Kiran

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