The speed of the top of a sliding ladder 
A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.

Is my solution wrong? How come I can't get the correct answer?
x=horizontal axis , y=vertical axis
Area=XY/2, when X=3,  A=6m^2, dx/dt=0.4m/s 
Y=2A/X 
dy/dx=-2A/x^2.....A=6,x=3
     =-12/9

dy/dt=dy/dx * dx/dt
     =-12/9 * 0.4
     =-0.533333


First we use Pythagorean Theorem
$$5^2=x^2+y^2 \implies y = \sqrt{25-x^2}.$$
Next we use that we can write $x=0.4t=\frac{2}{5}t$:
$$y=\sqrt{25-\left(\frac{2}{5}t\right)^2} = \sqrt{25-\frac{4}{25}t^2}.$$
Now we calculate the derivative (using the chain rule)
$$ \frac{dy}{dt}= \frac{\frac{d}{dt}\left(25-\frac{4}{25}t^2\right)}{2\sqrt{25-\frac{4}{25}t^2}} = \frac{-\frac{8}{25}t}{2\sqrt{25-\frac{4}{25}t^2}}.$$
This can be simplified further, but it isn't necessary. Now for the last step: evaluating the derivative at $t=7.5$ (which comes from solving $0.4t=3$).
$$\frac{dy}{dt}|_{t=7.5} =\frac{-\frac{8}{25}\cdot 7.5}{2\sqrt{25-\frac{4}{25}\cdot7.5^2}}=-0.3. $$
So the conclusion is that the speed, when the bottom is $3\text{m}$ from the wall, is $-0.3 \text{ m}/\text{s}$.
