Answer: \(x=-\dfrac32,-\dfrac72\); \(x=0.5\) gives a maximum and \(x=3\) gives a minimum; minimum surface area \(=81.4\text{ cm}^2\).

(a) Let

\(y=e^{3x}(2x+3)^6.\)

Use the product rule:

\(\frac{dy}{dx} =3e^{3x}(2x+3)^6+e^{3x}\cdot 12(2x+3)^5.\)

Factorise:

\(\frac{dy}{dx} =3e^{3x}(2x+3)^5\left[(2x+3)+4\right].\)

So

\(\frac{dy}{dx} =3e^{3x}(2x+3)^5(2x+7).\)

Stationary points occur when \(\frac{dy}{dx}=0\). Since \(e^{3x}\neq0\),

\((2x+3)^5(2x+7)=0.\)

Therefore

\(2x+3=0 \quad\text{or}\quad 2x+7=0.\)

Hence the \(x\)-coordinates are

\(x=-\frac32,\qquad x=-\frac72.\)

(b) Use the second derivative test.

At \(x=0.5\),

\(f''(0.5)=4(0.5)-7=2-7=-5.\)

Since \(f''(0.5)\lt 0\), the stationary point at \(x=0.5\) is a maximum.

At \(x=3\),

\(f''(3)=4(3)-7=12-7=5.\)

Since \(f''(3)\gt 0\), the stationary point at \(x=3\) is a minimum.

(c) The base is \(4x\) by \(x\), so the volume is

\(4x^2h.\)

Given that the volume is \(40\),

\(4x^2h=40.\)

Therefore

\(h=\frac{10}{x^2}.\)

The surface area is

\(S=2(4x\cdot x)+2(4xh)+2(xh).\)

So

\(S=8x^2+10xh.\)

Substitute \(h=\dfrac{10}{x^2}\):

\(S=8x^2+10x\left(\frac{10}{x^2}\right).\)

Thus

\(S=8x^2+\frac{100}{x}.\)

Differentiate:

\(\frac{dS}{dx}=16x-\frac{100}{x^2}.\)

At the minimum,

\(16x-\frac{100}{x^2}=0.\)

So

\(16x^3=100.\)

Hence

\(x^3=\frac{25}{4}.\)

Substituting this value of \(x\) into \(S=8x^2+\frac{100}{x}\) gives

\(S\approx81.4325.\)

Therefore the minimum surface area is

\(81.4\text{ cm}^2\)

to 3 significant figures.