University:
Whitman College
Course:
MATH-125 | Calculus I
Academic year:

2021
Views:

369

Pages:

1
Author:

Jerome Sherman

M125 Fall 2003 Quiz 5 SOLUTIONS Show all your work! Suppose you have a wire that is 10 meters long. You cut the wire, and with one piece bend it into the shape of a square. With the other piece, bend it into the shape of a circle. How should the wire be cut in order to have a minimum enclosed area? Let x be the length of the wire cut for the square, so that 10 − x is the length for the 10 − x circle. Since the circumference of a circle is 2πr, the radius of our circle will be r = . 2π The full area is then: 1 10 − x A(x) = x2 + π 16 2π 2 = 1 2 1 x + (10 − x)2 16 4π where 0 ≤ x ≤ 10. If x = 0, all the wire goes for the circle, and the area is 25 ≈ 7.96. π 1 If x = 10, all the wire goes for the square, and the area is 16 · 100 ≈ 6.25. Otherwise, we look for the critical points: 1 1 A0 (x) = x + (10 − x)(−1) = 0 8 2π 40 ≈ 5.601. π+4 Put this back into the area, and we get: so that the derivative is zero if x = 40 25 A = ≈ 3.5006 π+4 π+4 Looking back at the endpoints, this is the minimum value, obtained when x ≈ 5.6 and 10 − x ≈ 4.4 1

Calculus I Quiz 5 Solutions

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