An irrigation canal has the shape of a trapazoid. Let x =the bottom width, y =the water depth, and theta=the angle of inclination of the sides from the vertical.

Area=A=y(x+ytan(theta)) and Perimeter=P=x+2y/cos(theta)

It is known that the best design for a fixed inclination is found by minimizing P subject to
the constraint that the area A = Ao is constant. Show that this implies that
y^2 = Ao*cos(theta)/(2-Sin(theta))

Question

An irrigation canal has the shape of a trapazoid. Let x =the bottom width, y =the water depth, and theta=the angle of inclination of the sides from the vertical.

Area=A=y(x+ytan(theta)) and Perimeter=P=x+2y/cos(theta)

It is known that the best design for a fixed inclination is found by minimizing P subject to
the constraint that the area A = Ao is constant. Show that this implies that 
y^2 = Ao*cos(theta)/(2-Sin(theta))

anonymous · Answer

i finally understand how to find the area and the perimeter now Im stuck on this last part