Answer: (a) \(A=1\), \(B=14\); (c) \(\displaystyle \frac{5\sqrt3}{12}p\); (d) \(\displaystyle \frac{25\sqrt3}{24}\approx1.80\).

Answer: (a) \(A=1\), \(B=14\); (c) \(\displaystyle \frac{5\sqrt3}{12}p\); (d) \(\displaystyle \frac{25\sqrt3}{24}\approx1.80\).

Write

\(y=(2x+1)^{3/2}(x+5)^{-1}.\)

Differentiate using the product rule:

\(\frac{dy}{dx}=3(2x+1)^{1/2}(x+5)^{-1}-(2x+1)^{3/2}(x+5)^{-2}.\)

Take out the common factor \(\frac{(2x+1)^{1/2}}{(x+5)^2}\):

\(\frac{dy}{dx}=\frac{(2x+1)^{1/2}}{(x+5)^2}\left(3(x+5)-(2x+1)\right).\)

Simplify the bracket:

\(3(x+5)-(2x+1)=x+14.\)

Hence

\(\frac{dy}{dx}=\frac{(2x+1)^{1/2}}{(x+5)^2}(x+14),\)

so \(A=1\) and \(B=14\).

For \(x\ge0\), the factors \((2x+1)^{1/2}\), \((x+5)^2\), and \(x+14\) are all positive. Therefore \(\frac{dy}{dx}\) cannot be zero, so there are no stationary points.

When \(x=1\),

For a small increase \(p\),

\(\Delta y\approx \frac{dy}{dx}p=\frac{5\sqrt3}{12}p.\)

If \(\frac{dx}{dt}=2.5\) when \(x=1\), then

Thus the corresponding rate of change in \(y\) is approximately \(1.80\) units per second.