x: length of wire used for perimeter of square ; (10 - x): length of wire used for circumference of circle
The square then has side length x/4 and area x2/16.
The circle has circumference 10 - x = 2πr so r = (10 - x) / 2π . So area of circle = π (10 - x)2 / 4π2
A(x) = x2/16 + (10 - x)2 / 4π = x2/16 + x2/4π - 5x/π + 25/π = [(π + 4)/16π]⋅x2 - 5/π⋅x + 25/π ; for 0 ≤ x ≤ 10
This concave down parabola has a minimum at its vertex, x = 5/π / 2[(π + 4)/16π] = 40 / (π + 4).
Because the x-value of the minimum is closer to 10 than it is to 0, the max will be when x = 0.
Josh F.
tutor
Glad to help. Remember that the Extreme Value Theorem guarantees that this continuous function will have both an absolute max and an absolute min on the closed interval, [0 , 10], and the EVT also tells you where to look: these extrema will occur at critical values for f or at the endpoints of the interval (in this case the lefthand endpoint, 0). That result demonstrates that the way to enclose the maximal area with the wire is to shape the entire piece into a circle (a result that may even be intuitive).
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10/30/21
Huy N.
Thank you so much. I was able to find the minimum, but have no idea how to get the maximum. Your answer just save me.10/30/21