A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at its deepest point. If the pool is being filled at a rate of 0.8ft3/min, how fast is the water level rising whe

We can use the formula: V = lwh where V is the volume of the pool, l is the length, w is the width, and h is the height (depth) of the pool. First, we need to find the volume of the pool when the depth at the deepest point is 5 ft. We can use similar triangles to solve for the length and width at that depth: 9/5 = (40-x)/20 where x is the length of the shallow end (where the depth is 3 ft). Solving for x, we get x = 12.8 ft. So at a depth of 5 ft, the length is 27.2 ft and the width is still 20 ft. Now we can find the volume of the pool at that depth: V = (20)(27.2)(5 + 9 + 3)/3 = 1920 ft^3 Next, we need to find the rate of change of the volume with respect to time. This is given as 0.8 ft^3/min. Finally, we can calculate the rate of change of the water level (depth) at the deepest point using the formula: dV/dt = lwh(dh/dt) where dh/dt is the rate of change of the depth at the deepest point. Solving for dh/dt, we get: dh/dt = (dV/dt)/lw(h) Plugging in the values we know, we get: dh/dt = (0.8)/(20)(27.2)(9 - 5) = 0.00125 ft/min So the water level at the deepest point is rising at a rate of 0.00125 ft/min when the depth is 5 ft.