Area of Isosceles Triangle Problem

  • Thread starter Jimbo57
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  • #1
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Homework Statement


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?


Homework Equations





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?
 

Answers and Replies

  • #2
Dick
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Homework Helper
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Homework Statement


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?


Homework Equations





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?

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.
 
  • #3
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Dick, you've helped me on all of the questions that I've posted here. Thanks a bunch!
 

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