Optimization Problem

Question

A wire 5 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? What about maximize. Give the length of wire used for each:

Answer 1

x= the length of the part used to make an equilateral triangle.

1. Then each side of the triangle =x/3 
2. To determine the area of the triangle, we need to know how tall it is. Given the side length s, the height of an equilateral triangle is h = ½(√3s) = ½(√3 * x/3). Since, in our case s=x/3
3. Now we can compute the area of the equilateral triangle. The area is 1/2 base*height=1/2 (x/3 )* (½(√3 * x/3))=(x/6)(x√3/6)=(x²√3)/36
4. After cutting the amount of wire we need for the triangle, we will have x-5 meters left.
5. The radius of the circle used for the tv antenna with the remaining x-5 meters of wire is x-5/2π because circumference of the circle = x-5 = 2πr so r=x-5/2π
6. The area of the circle is π(x-5/2π)².
7. So now, the function we are maximizing is the sum of the areas of the equilateral triangle and the area of the circle. Hence the objective function is: f(x)=(x²√3)/36 + π(x-5/2π)²
8. To find extremum, f'(x), The first derivative of f(x) with respect to x will be set to 0. √3x/18 + x-5/2π=, x=(15(π√3-9)/(π²-27)), x=3.11604179.
9. Now we need to compute the second derivative f''(x) and validate that it is positive to ensure that the extremum x=3.11604179 is indeed a minimum. f''(x)= √3/18 + 12π > 0.

Therefore x=3.11604179 is the proper length we need to cut the wire to minimize the total area of both figures.

1. To maximize this function, we need to determine the maximum value of f(x)=(x²√3)/36 + π(x-5/2π)² over the domain x∈[0,5]. Since f(x) is quadratic and the leading coefficient of f(x) is positive, it increases monotonically on both sides of the minimum. So we only need to evaluate f(x) where x=0 and x=5. If you plot the f(x) graph, the maximum occurs at x=0. This means that the wire should be used to create the circle for the TV antenna. The maximum of f(x) = π(0-5/2π)²=25/4π =1.98943679

Answer 2

Easy to mess this up in the details:

Let piece of wire used for the antenna circle be L and border triangle 5 - L

A_C = πr² C = 2πr and A_C = π(C/(2π))² = C²/(4π) = L²/(4π)

A_T = s²sqrt(3)/4 P = 3s and A_T = P²sqrt(3)/36 = (5-L)²sqrt(3)/36

A_Total = A_C + A_T = L²/(4π) + (5-L)²sqrt(3)/36

Take dA/dL and equate to 0 to find extrema. The endpoint areas must be checked (A all circle, A all triangle) and compared to extremum in order to answer the questions. (The circle is the geometric figure that maximizes area given the perimeter of the figure).

Please consider a tutor. Take care.