Questions: A wire of length 100 cm is to be cut into two pieces. One piece is bent into a square, and the other is bent into a circle. (a) (2 points) How should the wire be cut to maximize the total area of the square and circle? (b) (2 points) How should the wire be cut to minimize the total area?

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Determine how to cut the wire to maximize the total area of the square and circle.
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Define the total area function.
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Express the areas of the square and circle in terms of the wire lengths.
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Let \( x \) be the length of the wire used for the square, and \( 100-x \) be the length used for the circle. The side length of the square is \( \frac{x}{4} \), so its area is \( \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \). The circumference of the circle is \( 100-x \), so its radius is \( \frac{100-x}{2\pi} \), and its area is \( \pi\left(\frac{100-x}{2\pi}\right)^2 = \frac{(100-x)^2}{4\pi} \). The total area is \( A(x) = \frac{x^2}{16} + \frac{(100-x)^2}{4\pi} \).
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Find the critical points of \( A(x) \).
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Take the derivative of \( A(x) \) and solve \( A'(x) = 0 \) to find critical points.
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The derivative is \( A'(x) = \frac{x}{8} - \frac{100-x}{2\pi} \). Setting \( A'(x) = 0 \), we get \( \frac{x}{8} = \frac{100-x}{2\pi} \), which simplifies to \( 2\pi x = 800 - 8x \). Solving for \( x \), we find \( x = \frac{800}{2\pi + 8} = \frac{400}{\pi + 4} \approx 56.01 \).
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Evaluate \( A(x) \) at the endpoints and the critical point.
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Compare \( A(x) \) at \( x=0 \), \( x=100 \), and \( x=\frac{400}{\pi+4} \).
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At \( x=0 \), \( A(0) = \frac{100^2}{4\pi} = \frac{2500}{\pi} \approx 795.77 \).
At \( x=100 \), \( A(100) = \frac{100^2}{16} = 625 \).
At \( x=\frac{400}{\pi+4} \), \( A\left(\frac{400}{\pi+4}\right) = \frac{2500}{\pi+4} \approx 350.06 \).
The maximum area occurs at \( x=0 \), and the minimum area occurs at \( x=\frac{400}{\pi+4} \).
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To maximize the total area, use 0 cm for the square and 100 cm for the circle.

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Determine how to cut the wire to minimize the total area of the square and circle.
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Identify the wire lengths for minimizing the total area.
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Use the critical point \( x=\frac{400}{\pi+4} \) to minimize the area.
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At \( x=\frac{400}{\pi+4} \approx 56.01 \), the wire is divided into approximately 56.01 cm for the square and \( 100 - \frac{400}{\pi+4} \approx 43.99 \) cm for the circle. This configuration minimizes the total area.
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To minimize the total area, use approximately 56.01 cm for the square and 43.99 cm for the circle.
☺ To maximize the area, use 0 cm for the square and 100 cm for the circle. To minimize the area, use $\frac{400}{\pi+4} \approx 56.01$ cm for the square and $100 - \frac{400}{\pi+4} \approx 43.99$ cm for the circle.