Results 1 to 3 of 3

Thread: Cutting Wire into Two Pieces

  1. #1
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Cutting Wire into Two Pieces

    A wire 10 meters long is to be cut into two pieces. One piece will be shaped as an equilateral triangle and the other piece will be shaped as a circle.

    (a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle.

    (b) What is the domain of the area A?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, magentarita!

    A wire 10 meters long is to be cut into two pieces. One piece will be bent
    into an equilateral triangle and the other piece will be bent into a circle.

    (a) Express the total area $\displaystyle A$ enclosed by the pieces of wire as a function
    of the length $\displaystyle x$ of a side of the equilateral triangle.
    Let $\displaystyle x$ = side of the triangle.
    The area of an equilateral triangle of side $\displaystyle x$ is: .$\displaystyle A_t \:=\:\frac{\sqrt{3}}{4}x^2$


    The triangle uses $\displaystyle 3x$ of the wire.
    This leaves $\displaystyle 10-3x$ for the circumference of the circle.

    Circumference formula: .$\displaystyle C \:=\:2\pi r$

    So we have: .$\displaystyle 10-3x \:=\:2\pi r \quad\Rightarrow\quad r \:=\:\frac{10-3x}{2\pi} $

    Area formula: .$\displaystyle A \:=\:\pi r^2$

    So we have: .$\displaystyle A_c \;=\;\pi\left(\frac{10-3x}{2\pi}\right)^2 \quad\Rightarrow\quad A_c\;=\;\frac{(10-3x)^2}{4\pi}$


    Therefore: . $\displaystyle \boxed{A \;=\;\frac{\sqrt{3}}{4}x^2 + \frac{(10-3x)^2}{4\pi}} $



    (b) What is the domain of $\displaystyle A$ ?

    Domain: .$\displaystyle 0 \,\leq\,x\,\leq\,3\tfrac{1}{3}$


    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Soroban....

    Quote Originally Posted by Soroban View Post
    Hello, magentarita!

    Let $\displaystyle x$ = side of the triangle.
    The area of an equilateral triangle of side $\displaystyle x$ is: .$\displaystyle A_t \:=\:\frac{\sqrt{3}}{4}x^2$


    The triangle uses $\displaystyle 3x$ of the wire.
    This leaves $\displaystyle 10-3x$ for the circumference of the circle.

    Circumference formula: .$\displaystyle C \:=\:2\pi r$

    So we have: .$\displaystyle 10-3x \:=\:2\pi r \quad\Rightarrow\quad r \:=\:\frac{10-3x}{2\pi} $

    Area formula: .$\displaystyle A \:=\:\pi r^2$

    So we have: .$\displaystyle A_c \;=\;\pi\left(\frac{10-3x}{2\pi}\right)^2 \quad\Rightarrow\quad A_c\;=\;\frac{(10-3x)^2}{4\pi}$


    Therefore: . $\displaystyle \boxed{A \;=\;\frac{\sqrt{3}}{4}x^2 + \frac{(10-3x)^2}{4\pi}} $



    Domain: .$\displaystyle 0 \,\leq\,x\,\leq\,3\tfrac{1}{3}$

    You are fantastic. I love your replies. I am going to take calculus 1 soon. Are you interested in helping me as I go through calculus 1?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Mar 23rd 2011, 05:59 AM
  2. cutting the cube
    Posted in the Math Puzzles Forum
    Replies: 5
    Last Post: Oct 15th 2010, 10:41 PM
  3. Replies: 2
    Last Post: Apr 27th 2010, 06:14 AM
  4. Optimization: A Wire is to be cut into Two Pieces.
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: Mar 28th 2010, 05:31 PM
  5. Pieces of Wire
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jun 13th 2007, 09:36 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum