Artist also needs Mathematics!

Relevant wiki: Arithmetic Mean - Geometric Mean

Let the dimensions of the picture be $x \times y$ . Therefore $xy=72$ . Then the area of the canvas is:

$\begin{aligned} ab & = (x+8)(y+4) & \small \color{#3D99F6} \text{For }a \text{ and }x \text{ being heights.} \\ & = {\color{#3D99F6}xy} + 4x + 8y + 32 & \small \color{#3D99F6} \text{Note that }xy=72 \\ & = {\color{#3D99F6}72} + 4x + 8y + 32 \\ & = 4{\color{#3D99F6}(x + 2y)} + 104 & \small \color{#3D99F6} \text{By AM-GM inequality:} \\ & \ge 4{\color{#3D99F6}(24)} + 104 & \small \color{#3D99F6} x+2y \ge 2 \sqrt{2xy} = 2\sqrt{144} = 24 \\ & = 96 + 104 & \small \color{#3D99F6} \text{Equality when: }x=12, y=6 \\ & = \boxed{200} \end{aligned}$

Let the width of the image be $w$ and therefore the height $\frac {72}w$ . Then, the dimensions of the canvas are $(2+w+2) \times (4+\frac{72}w+4) = 8w +\frac{288}{w} +104$ .

To minimize this, we take the derivative and set it equal to zero.

$\frac d {dw} = 8 - \frac {288} {w^2} \stackrel{set}= 0 \Leftrightarrow 8w^2 = 288 \Leftrightarrow w = \pm 6$

We can ignore $w=-6$ because we're talking about sizes.

Then, we plug in $w=6$ into our first equation to get $(2+6+2) \times (4 + \frac{72}6 + 4) = 10 \times 20 = \boxed{200}$