# Application on Differentiation

• Lord Dark
In summary, the problem asks for the best way to cut a 20 cm wire into two pieces in order to maximize or minimize the total area enclosed by an equatorial triangle and a circle. The solution involves using x as the length of the circle and (20-x) as the triangle length, finding the area of each shape, differentiating the sum of the areas, and solving for the value of x that gives a maximum or minimum area. An easier approach is suggested, using the cosine function to calculate the height of the triangle, and checking the solution with a graph.

## Homework Statement

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 ?

## 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 ...

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.