Answer: (i) \(A=\left(\dfrac32,36\right)\). (ii) Area \(=12\).

(i) Differentiate:

\(y=16x+27x^{-2},\)

so

\(\frac{dy}{dx}=16-54x^{-3} = 16-\frac{54}{x^3}.\)

At the minimum,

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

Therefore

\(16-\frac{54}{x^3}=0.\)

So

\(16x^3=54, \qquad x^3=\frac{27}{8}.\)

Hence

\(x=\frac32.\)

Now find \(y\):

\(y=16\left(\frac32\right)+\frac{27}{\left(\frac32\right)^2}.\)

This gives

\(y=24+\frac{27}{9/4}=24+12=36.\)

So

\(A=\left(\frac32,36\right).\)

(ii) First find the coordinates of \(P\) and \(Q\).

When \(x=1\),

\(y=16+\frac{27}{1}=43.\)

So \(P=(1,43)\).

When \(x=3\),

\(y=48+\frac{27}{9}=51.\)

So \(Q=(3,51)\).

The gradient of \(PQ\) is

\(\frac{51-43}{3-1}=4.\)

Using \(P=(1,43)\), the equation of \(PQ\) is

\(y-43=4(x-1),\)

so

\(y=4x+39.\)

The required area is

\(\int_1^3\left(4x+39-\left(16x+\frac{27}{x^2}\right)\right)\,dx.\)

Simplify the integrand:

\(4x+39-16x-\frac{27}{x^2} = -12x+39-\frac{27}{x^2}.\)

Integrate:

\(\int\left(-12x+39-27x^{-2}\right)\,dx = -6x^2+39x+\frac{27}{x}.\)

Now apply the limits:

\(\left[-6x^2+39x+\frac{27}{x}\right]_1^3.\)

At \(x=3\),

\(-6(9)+39(3)+\frac{27}{3} = -54+117+9=72.\)

At \(x=1\),

\(-6+39+27=60.\)

So the required area is

\(72-60=12.\)