Represent real-world situations a rectangular piece of sheet metal is rolled and riveted to form a circular tube that is open at both ends, as shown. the sheet metal has a perimeter of 36 inches. each of the two sides of the rectangle that form the two ends of the tube has a length of x inches, and the tube has a circumference of x - 1 inches because an overlap of 1 inch is needed for the rivets. write a volume function for the tube in terms of x. then, to the nearest tenth, find the value of x that maximizes the volume of the tube. answer

12.3 Volume function: v(x) = ((18-x)(x-1)^2)/(4pi) Since the perimeter of the piece of sheet metal is 36, the height of the tube created will be 36/2 - x = 18-x. The volume of the tube will be the area of the cross section multiplied by the height. The area of the cross section will be pi r^2 and r will be (x-1)/(2pi). So the volume of the tube is v(x) = (18-x)pi((x-1)/(2pi))^2 v(x) = (18-x)pi((x-1)^2/(4pi^2)) v(x) = ((18-x)(x-1)^2)/(4pi) The maximum volume will happen when the value of the first derivative is zero. So calculate the first derivative: v'(x) = (x-1)(3x - 37) / (4pi) Convert to quadratic equation. (3x^2 - 40x + 37)/(4pi) = 0 3/(4pi)x^2 - (10/pi)x + 37/(4pi) = 0 Now calculate the roots using the quadratic formula with a = 3/(4pi), b = -10/pi, and c = 37/(4pi) The roots occur at x = 1 and x = 12 1/3. There are the points where the slope of the volume equation is zero. The root of 1 happens just as the volume of the tube is 0. So the root of 12 1/3 is the value you want where the volume of the tube is maximized. So the answer to the nearest tenth is 12.3

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