I am having issues getting the correct answer for this question. I'll provide my attempted solution, as well as the book's answer.
Question
A swimming pool is 25 ft wide, 40 ft long, 3 ft deep at one end and 9 ft deep at the other, the bottom being an inclined plane. If water is pumped into the pool at the rate of 10 ft$^3$/min, how fast is the water level rising when it is 4 ft deep at the deep end?
Solution (Attempt)
Around the 4 ft mark the volume is given by a triangular prism, i.e.
$$
V = \frac{1}{2}(25xy)
$$
where $y$ is the depth of the water, and $x$ is the length as in the picture. Thus
$$
\frac{dV}{dt} = \frac{25}{2}\left(y\frac{dx}{dt} + x\frac{dy}{dt}\right)
$$
Furthermore, by similar triangles we get that
$$
\frac{y}{x} = \frac{9}{40}
$$
that is
$$
x = \frac{40}{9}y
$$
so that
$$
\frac{dx}{dt} = \frac{40}{9}\frac{dy}{dt}
$$
Also by similar triangles we get that when $y=4$ then $x=\frac{160}{9}$.
Inserting the values for $x$ and $y$, as well as making use of the equation for $\frac{dx}{dt}$ into the equation for the rate of change for the volume, and using the fact that the change in volume should be 10, we get
$$
10 = \frac{25}{2}\left(4\frac{40}{9}\frac{dy}{dt} + \frac{160}{9}\frac{dy}{dt}\right)
$$
i.e.
$$
\frac{dy}{dt} = \frac{9}{400}
$$
However, the answer in the back of the book reads $\frac{3}{200}$.
Best Answer
I incorrectly thought the left side was 9 ft, and not 6 ft. Changing this solves the problem.