Question

The sum of the circumference of a circle and the perimeter of a square is 16. find the dimensions of the circle and square that produce a minimum total area. (let x be the length of a side of the square and r be the radius of the circle.)

Answer 1

Let r =  the radius of the circle.
Let x =  the length of a side of the square.

The sum of the circumference of the circle and the perimeter of the square is 16.
Therefore
2πr + 4x = 16
That is
$r = \frac{2}{ \pi } (4-x)$          (1)

The combined area of the circle and the square is
$A = \pi r^{2} + x^{2}$            (2)

From (1), obtain
$A = \pi . \frac{4}{ \pi ^{2}}(4-x)^{2} + x^{2} = \frac{4}{ \pi } (4-x)^{2}+x^{2}$

In order for A to be minimum, A' = 0.
That is,
$- \frac{8}{ \pi } (4-x)+2x=0 \\x(2+ \frac{8}{ \pi } ) = \frac{32}{ \pi } \\ x =2.2404$
The second derivative should be positive in order for A to be minimum.
A''  = 8/π + 2 > 0 , so the condition is satisfied.
The graph shown below confirms this.

From (1), obtain
r = (2/π)*(4 - 2.2404) = 1.1202

Answer:
r = 1.12 and x = 2.24