Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
MAX FLOW A company is constructing guttering to carry water. The cross section of the guttering is below: x x w 2x Each side is the same length and the angle between each side and the horizontal are equal. That is, the cross section is symmetrical about the vertical line through the mid point of its base. Denote the width of the material used to make the guttering as w. Denote the length of each side as x Denote the ration of w as k. x Denote the angle of inclination of the sides as . TASK A If 90 , find the value of k to maximise the volume of water that the guttering can transport. When 90 , the cross section x w-2x and we are to maximize the area of S = x*(w-2x) or S= xw – 2x^2 take the first derivative of S, we get w-4x=0 or x=w/4 then k = w/x = w/ (w/4) = 4 this yields the same result if you write: S = xw – 2x^2 = – 2(x^2 – xw/2) = – 2(x^2 – 2(w/4)*x +(w/4)^2) + 2(w/4)^2 = – 2(x – (w/4))^2 + 2(w/4)^2 Which is maximized when x = w/4 And the volume of water that the guttering can transport is maximised. TASK B If k 2 , find the value of to maximise the volume of water that the guttering can transport.. If k 2 , x = w/2, the bottom will be 0, then the cross section is a h x Therefore, the length of a=x*cos and height is h=x*sin Now we are to maximize the area of S = (x*cos )(x*sin ) or S= x^2 * (sin *cos ) which is maximized when sin =cos i.e. = 45 degree and sin =cos = SQRT(2)/2 And the volume of water that the guttering can transport is maximised. TASK C If 60 , find the value of k to maximise the volume of water that the guttering can transport. c h x w-2x If 60 , the height of the cross section is h=x*sin(60)= sqrt(3)x /2 Then length c =x*cos(30)= x/2 And the bottom is b=w-2x, while the width of the opening is then a=2c+b= 2* x/2 + w-2x = w-x now the area of the cross section is S=h(a+b)/2 = sqrt(3)x /2 * (w-x + w-2x) /2 = sqrt(3)/4 * x(2w-3x) take the first derivative of S, we get sqrt(3)/4 * (2w-6x) =0 or x=w/3 then k = w/x = w/ (w/3) = 3 And the volume of water that the guttering can transport is maximised. NOTES Use calculus to find all values of all the tasks listed above. Show all workings and use both graphs and numbers to denote workings along with diagrams of the guttering where necessary.