Answer: stationary point \((2\sqrt2,8)\); \(-3x(16-x^2)^{1/2}\); area \(13.2\).

(a) The curve is

\(y=x(16-x^2)^{1/2}.\)

Using the product rule,

\(\frac{dy}{dx}=(16-x^2)^{1/2} x\cdot\frac12(16-x^2)^{-1/2}(-2x).\)

So

\(\frac{dy}{dx}=(16-x^2)^{1/2}-\frac{x^2}{(16-x^2)^{1/2}}.\)

At a stationary point,

\(\frac{dy}{dx}=0.\)

Therefore

\((16-x^2)^{1/2}=\frac{x^2}{(16-x^2)^{1/2}}.\)

Multiplying by \((16-x^2)^{1/2}\),

\(16-x^2=x^2.\)

Thus

\(x^2=8.\)

Since \(0\leq x\leq4\),

\(x=2\sqrt2.\)

Then

\(y=2\sqrt2\sqrt{16-8}=2\sqrt2\sqrt8=8.\)

So the stationary point is

\((2\sqrt2,8).\)

(b)(i) Differentiate using the chain rule:

\(\frac{d}{dx}(16-x^2)^{3/2} =\frac32(16-x^2)^{1/2}(-2x).\)

Hence

\(\frac{d}{dx}(16-x^2)^{3/2} =-3x(16-x^2)^{1/2}.\)

(b)(ii) The required area is

\(\int_1^3 x(16-x^2)^{1/2}\,dx.\)

From part (b)(i),

\(\int x(16-x^2)^{1/2}\,dx =-\frac13(16-x^2)^{3/2}.\)

Therefore the area is

\(\left[-\frac13(16-x^2)^{3/2}\right]_1^3.\)

So

\(\text{area} =-\frac13(7)^{3/2}+\frac13(15)^{3/2} =\frac{15\sqrt{15}-7\sqrt7}{3}.\)

This is

\(13.2\)

to 3 significant figures.