SOLUTION: A 10m rope is used to enclose a square area with side x and a circle area with radius r. How much rope should be used for the square and how much for the circle if the total area e

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 Question 451174: A 10m rope is used to enclose a square area with side x and a circle area with radius r. How much rope should be used for the square and how much for the circle if the total area enclosed by the two shapes is to be a max? Plot total area enclosed by two shapes as a function of x, where x varies from .05 to 5m, and please show graphically that the area reaches a max at x=1.4m. I think the x at the max total area = 10/(pi+4). I tried using matlab, any matlab commands would help! Answer by stanbon(57377)   (Show Source): You can put this solution on YOUR website!A 10m rope is used to enclose a square area with side x and a circle area with radius r. How much rope should be used for the square and how much for the circle if the total area enclosed by the two shapes is to be a max? Plot total area enclosed by two shapes as a function of x, where x varies from .05 to 5m, and please show graphically that the area reaches a max at x=1.4m. I think the x at the max total area = 10/(pi+4). --- Perimeter of square = 4x meters Perimeter of circle = 2(pi)r ----------- Equations: Perimeters:::: 4x + 2(pi)r = 10 meters Areas = [x^2 + (pi)r^2] sq. meters ------ Modify the 1st equation: 2x+(pi)r = 5 r = (5-2x)/pi ------ Substitute for "r and solve for "x": Areas = x^2 + (pi)[(5-2x)/pi]^2 Areas = x^2+(5-2x)^2/pi Areas = x^2 + 25/pi - 20x/pi + 4x^2/pi Areas = [(4/pi)+1]x^2 - (20/pi)x + (25/pi) Area = ((4 + pi)/pi)x^2 - (20/pi)x + (25/pi) ---- x varies from .05 to 5m: Find the max area on the interval [0.05,5] A(0.05) = 7.6573 A(5) = 154.67 ---- Max area on [0.05,5] = 154 sq. meters ================= Cheers, Stan H.