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Area of Isosceles Triangle Problem

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data
    An Isosceles triangle has two equal sides of length 10cm. Let x be the angle between the two equal sides.

    a. Express the area A of the triangle as a function of x in radians.

    b. Suppose that x is increasing at the rate of 10 degrees per minute. How fast is A changing at the instant x = pi/3? At what value of x will the triangle have a maximum area?


    2. Relevant equations



    3. The attempt at a solution

    For part A, I know the answer is 50sin(x) using the double angle formula

    for B)
    dx/dt=10degrees=0.1745rad

    da/dt=50cos(x)(dx/dt)
    =50cos(pi/3)(0.1745)
    =4.36cm^2/min

    For max area, using optimization techniques:

    A'=50cosx >0 for 0<x<pi/2 and 3pi/2<x<2pi (since dimensions can't be negative)
    A' <0 for pi/2<x<3pi/2
    A'= 0 for x= pi/2 and 3pi/2
    Using closed interval method:
    A(pi/2)=50
    A(3pi/2)=-50, therefore A will be max when x=pi/2

    I put this in another thread from about a year ago but the information was scrambled and I didn't receive any feedback, anyone mind taking a look at this to see if it's correct?
     
  2. jcsd
  3. Jan 19, 2012 #2

    Dick

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    That looks ok. Except I would say the correct range for x should 0<=x<=pi. Just because all of the angles of a triangle should be in that range.
     
  4. Jan 19, 2012 #3
    Dick, you've helped me on all of the questions that I've posted here. Thanks a bunch!
     
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