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 February 26th, 2008, 09:26 PM #1 Newbie   Joined: Feb 2008 Posts: 2 Thanks: 0 find the dimension for isosceles trapezoid question: a) An isosceles trapezoid drainage gutter is to be made so that the angles at A and B , in the cross section ABCD are each 120 degree. If the 5m long sheet of metal that has to be bent to form the open topped gutter has a width of 60cm, then find the dimensions so that the cross - sectional area will be a maximum. b) calculate the max volumn of water the can be held by this gutter
 February 27th, 2008, 12:45 PM #2 Member   Joined: Feb 2008 From: Dayton, OH, USA Posts: 33 Thanks: 0 This is a good problem. I whipped up a simple diagram to go along with my solution: Given that two of the angles must always be 120 deg, by virtue of it being a quadrilateral and an isosceles trapezoid, the other two angles can be easily found: Let alpha = angle A = angle B = 120 deg and let beta = angle C = angle D So 2 * alpha + 2 * beta = 360 deg Thus beta = (360 - 2* alpha) / 2 = 60 deg Given the requirements of the geometry there are really only two variables here: the length of the base AB and the length of the two equal side DA and BC. Let us call these two variables Lb and Ls respectively. Furthermore since we know the total width of the sheet, let us call this L which is 60 cm or 0.6 m, we can relate these two variables: Lb + 2*Ls = L Thus Lb = L - 2*Ls So now we are effectively down to only one variable and we now need to find the cross-sectional area in terms of this one variable. Since generally the area of an isosceles trapezoid is given in terms of its height and two bases let us derive this ourselves in terms of our variables. To do think let us think of the isosceles trapezoid as a rectangle in the center with a width Lb and height h. Attached at either end of this rectangle are two equal (but reflected) right triangles with height h and hypotenuse Ls. The height and width of the triangle can be found with trig: h = Ls * sin(beta) w = Ls * cos(beta) The area is then A = Arectangle + 2 * Atriangles A = Lb * h + 2 * 1/2 * h * w If we substitute for Lb, w, and h in terms of Ls we obtain area as a function of Ls: A(Ls) = (L - 2*Ls)*Ls*sin(beta) + Ls*sin(beta)*Ls*cos(beta) A(Ls) = sin(beta)*[cos(beta) - 2]*Ls^2 + L*sin(beta)*Ls To maximize this function we set the derivative to 0: A'(Ls) = 2*sin(beta)*[cos(beta) - 2]*Ls + L*sin(beta) = 0 Solving for Ls we get: Ls = -L / (2[cos(beta) - 2]) Plugging in the numbers we get Ls = 20 cm and plugging that into our previous equations we get: Lb = Ls = 20 cm and A = 520 cm^2 = 0.052 m^2 To find volume just multiple the cross sectional area by the length of the sheet: V = A*l = 259808 cm^3 = 0.26 m^3

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isosceles trapezoid with angles of 120 and 60

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