Calculus Problem

Let $x$ be the perimeter of the hexagon. Then $30-x$ is the perimeter of the square.

The hexagon has side length $\frac{x}{6}.$ It can be divided into 6 congruent equilateral triangles, and one can calculate the area of each of these triangles by using 30-60-90 triangle side length relationships. The area of the hexagon is $\frac{x^2 \sqrt{3}}{24}.$

The square has side length $\frac{30-x}{4}.$ It has an area of $\frac{(30-x)^2}{16}.$

The total area of square and hexagon is:

$A=\frac{x^2 \sqrt{3}}{24}+\frac{(30-x)^2}{16}.$

Take the derivative and set it to 0 to find the local extrema.

$0=\frac{x\sqrt{3}}{12}-\frac{30-x}{8}$

Solving this equation gives $x=60\sqrt{3}-90.$

Substituting this value of $x$ into the area equation gives $A\approx 30.144.$ It's also necessary to test the values of $x=0$ and $x=30.$ These values give $A=56.25$ and $A\approx 64.952,$ respectively. The minimum area is $\boxed{30.144}.$