Hello, ((ECHO))!

I get the same answer is behemoth and Jhevon . . .

A swimming pool is 22 ft wide and 60 ft long.

It is 2 ft deep at the shallow end, and 10 ft deep at the other end.

If the pool is being filled with water at a rate of 0.8 ft³/min,

how fast is the water level rising when the depth at the deepest point is 7 ft? Code:

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Since $\displaystyle \Delta BAE \sim \Delta DCE$, we have: .$\displaystyle \frac{x}{h} = \frac{60}{8}\quad\Rightarrow\quad x = \frac{15}{2}h$ .**[1]**

. . Hence: .$\displaystyle \frac{dx}{dt} = \frac{15}{2}\!\cdot\!\frac{dh}{dt}$ . **[2]**

The volume of water is: .$\displaystyle V \:=\:\frac{1}{2}xh(22) \:=\:11xh$

. . Then: .$\displaystyle \frac{dV}{dt}\;=\;11x\!\cdot\frac{dh}{dt} + 11h\!\cdot\!\frac{dx}{dt} $ . **[3]**

When $\displaystyle h = 7$, [1] gives us: .$\displaystyle x \:=\:\frac{15}{2}(7) \:=\:\frac{105}{2}$

Substitute into [3]: .$\displaystyle 0.8 \;=\;11\left(\frac{105}{2}\right)\left(\frac{dh}{d t}\right) + 11(7)\left(\frac{15}{2}\!\cdot\!\frac{dh}{dt}\righ t) $

Hence: .$\displaystyle 1155\left(\frac{dh}{dt}\right) \:=\:0.8\quad\Rightarrow\quad \frac{dh}{dt}\:=\:\frac{0.8}{1155}$

Therefore: .$\displaystyle \frac{dh}{dt} \:=\:0.000692641

$ ft/min.