Place four pennies on a table such that each touches at least one other.
For what arrangement will the area of the convex hull of this shape be maximized?
For simplicity let the pennies have unit radius.
Arrange the four pennies in a U-shape like in my first post but allow the width of the U to vary, while keeping the shape symmetrical.
Consider the isosceles trapezoid formed by the centers of the pennies. Denote its perimeter as Pt and area as At. Then the area of the convex hull of the pennies is 1^2*pi+1*Pt+At, or simply pi+Pt+At.
The shorter base and the two diagonal sides of the trapezoid are length 2. Let 2x be the length of the longer base. Then Pt = 6+2x and At = sqrt[(1+x)^3*(3-x)]. Because an isosceles trapezoid is a cyclic quadrilateral, At can be derived from Area = sqrt[(s-a)*(s-b)*(s-c)*(s-d)] where s is the semiperimeter and the sides are a,b,c,d.
Then the total area is:
pi + (6+2x) + sqrt[(1+x)^3*(3-x)]
This is valid for x=1 to 3. At x=1, the pennies are in a square (area = pi+12). At x=3 they are in a straight line (area = pi+12).
The first derivative is:
2 + [3*(1+x)^2*(3-x)-(1+x)^3]/(2*sqrt[(1+x)^3*(3-x)])
Equating this to zero and solving yields a quartic equation:
x^4 - 2x^3 - 2x^2 + 2x + 1 = 0
This has roots x=-1, x=1, x=1-sqrt[2] and x=1+sqrt[2]. The only one that is inside the range x=1 to 3 is x=1+sqrt[2].
Plugging x=1+sqrt(2) back into the area formula yields a maximum area of pi + 10 + 4*sqrt[2]. For this area, the distance of the long side of the trapezoid (the centers of the two furthest apart pennies) is 2+2*sqrt[2].
A possible connected maximum
