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Application on Differentiation

  • Thread starter Lord Dark
  • Start date
  • #1
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Homework Statement


could someone please help me to answer the following problem:

Suppose a wire 20 cm long is to be cut into two pieces. One piece is to be bent in the shape of an equatorial triangle and the other in the shape of a circle. How should the wire be cut so as to:
a) maximize the total area enclosed by the shapes ?

b)minimize the total area enclosed by the shapes ?

Homework Equations





The Attempt at a Solution


i applied x as the circle length and (20-x) as the triangle length ,,
we know that x=(2pi)r > r=x/2pi ,, A(c)=pi*(x/2pi)^2 ,,
A(t)=.5*(20-x)/3*sqrt(((20-x)/3)^2-((20-x)/6)^2)
A(c+t)= A(c) + A(t) ... then differentiate ,, solve for A`(c+t)=0 and i'll get what ?? maximize or minimize and how to get the other one ?? ,, and is there another way ?? (easier one) because it's hard to solve for A`(c+t) the equation is too long ...
 

Answers and Replies

  • #2
288
0
For the area of the triangle calculate the height by using the following:

[tex]h=\frac{1}{2}base*\cos(60)[/tex]

rather than using the Pythagorean Theorem. It will lead to a cleaner expression that will be much easier to differentiate. Take the second derivative of your expression to determine if it's a maximum or minimum.
 
  • #3
121
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lol ,, thanks :)
 
  • #4
121
0
ok ,, i got A(c+t)= x^2/4pi + (sqrt(3)/36) * (20-x)^2
i differentiate and solved for 0 i got x = 7.53583283 and got:
the maximum A(20) and minimum A(x) ,, is it right ??
 
  • #5
288
0
If you have a graphing calculator, graph it and find the minimum. Otherwise, graph it on paper to see if your value is correct. Should be an easy check.
 

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