A wire 100 m is to be cut into two pieces. One piece is bent into a square and the other into a circle. Where should the cut be made if the sum of the two areas is to be a maximum?
Answer (2)
The optimal cut to maximize the total area is at 0 m, meaning the entire 100 m of wire should be used to form the circle and no cut should be made for the square. This results in the maximum area being achieved. Hence, no cut should be made.
;
Define x as the length of wire for the square, leaving 100 − x for the circle.
Express the total area A as the sum of the square's and circle's areas: A = 16 x 2 + 4 π ( 100 − x ) 2 .
Find the critical point by setting d x d A = 0 , resulting in x = π + 4 400 ≈ 56.01 .
Evaluate A at endpoints x = 0 , x = 100 and the critical point; the maximum area occurs at x = 0 , meaning no cut should be made and all wire should be used for the circle. The answer is 0
Explanation
Problem Analysis We are given a 100 m wire that needs to be cut into two pieces. One piece will be bent into a square, and the other into a circle. Our goal is to determine where to cut the wire to maximize the total area enclosed by the square and the circle.
Define Variables Let x be the length of the wire used for the square. Then, the remaining length, 100 − x , will be used for the circle.
Square Area The side length of the square is s = 4 x . The area of the square is A s = s 2 = ( 4 x ) 2 = 16 x 2 .
Circle Area The radius of the circle is r = 2 π 100 − x . The area of the circle is A c = π r 2 = π ( 2 π 100 − x ) 2 = 4 π ( 100 − x ) 2 .
Total Area The total area is the sum of the square's area and the circle's area: A = A s + A c = 16 x 2 + 4 π ( 100 − x ) 2 .
Derivative of Total Area To find the maximum area, we need to find the critical points by taking the derivative of A with respect to x and setting it to zero: d x d A = 16 2 x + 4 π 2 ( 100 − x ) ( − 1 ) = 8 x − 2 π 100 − x .
Critical Point Setting the derivative to zero: 8 x − 2 π 100 − x = 0 . Solving for x : 8 x = 2 π 100 − x ⟹ π x = 400 − 4 x ⟹ x ( π + 4 ) = 400 ⟹ x = π + 4 400 ≈ 56.01 .
Endpoint and Critical Point Evaluation Now we need to check the endpoints ( x = 0 and x = 100 ) and the critical point ( x = π + 4 400 ) to determine the maximum area.
At x = 0 : A = 16 0 2 + 4 π ( 100 − 0 ) 2 = 4 π 10000 ≈ 795.77 .
At x = 100 : A = 16 10 0 2 + 4 π ( 100 − 100 ) 2 = 16 10000 = 625 .
At x = π + 4 400 ≈ 56.01 : A = 16 ( 56.01 ) 2 + 4 π ( 100 − 56.01 ) 2 ≈ 350.06 .
Comparing the values, we see that the maximum area occurs when x = 0 .
Conclusion The maximum area occurs when x = 0 , meaning all the wire should be used to form the circle. Therefore, no cut should be made.
Examples
In architecture, determining how to allocate a fixed perimeter to maximize enclosed area is crucial for optimizing space usage in building designs. For example, when designing a garden with a fixed fence length, deciding whether to enclose a circular or square area (or a combination of both) can significantly impact the garden's size. This problem demonstrates how calculus can be applied to make informed decisions about resource allocation to achieve maximum efficiency.
