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A water trough is 10 m long and a cross-section has the shape

of an isosceles trapezoid that is 30 cm wide at the bottom,

80 cm wide at the top, and has height 50 cm. If the trough is

being filled with water at the rate of 0.2 m^3͞/min, how fast is the

water level rising when the water is 30 cm deep?

I broke up the trapezoid into 2 triangles and a rectangle so the area should be

A = 2*Triangle + Rectangle

A = (b*h)+(50*h)

I can look at b in terms of h by using similar triangles. Each triangle has side a, of 25 side b, of 50. So the triangle formed by the water level must be side a 15 side b of 30. So b is h/2

A = (h^2*1/2) + (50*h)

Converting this into volume should be easy, just multiply the equation A, by 1000.

dA/dt = 2/2*h*dh/dt + 50*dh/dt

Convert dA/dt into dV/dt;

1000*dA/dt = dh/dt(h+50)*1000

Since 1000*dA/dt I can make a substitution

dV/dt = dh/dt(h+50)*1000

I have a book answer of 10/3. Could someone be so kind as to point out the errors? Thanks!