Application on Differentiation

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To maximize the total area enclosed by a wire cut into two pieces for a triangle and a circle, the wire length designated for the circle is represented as x, while the remaining length for the triangle is (20-x). The area for the circle is calculated using A(c) = π*(x/2π)², and the area for the triangle is derived from its height using h = (1/2)base*cos(60). After differentiating the total area A(c+t) and setting it to zero, the critical point found is x = 7.54, indicating a maximum area. For minimizing the area, the same differentiation process applies, and graphing can confirm the results. The discussion emphasizes the need for a cleaner expression for easier differentiation and verification through graphing.
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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 ...
 
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For the area of the triangle calculate the height by using the following:

h=\frac{1}{2}base*\cos(60)

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.
 
lol ,, thanks :)
 
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 ??
 
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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