you have a piece of wire of length 10 ft from which you construct a circle and/or a square in such a way that the total enclosed area is maximal. what radius does the circle have and what side length does the square have to maximize the enclosed area?

Question

campbell_st · Answer

Let the perimeter of the square me x, then  the circle has a circumference of 10 - x

the. 

the length of 1 side of the square is x/4 and area = x^2/16

the circle 

$$10 - x = 2 \pi r$$

so $$(10 -x)/2\pi = r$$

Area of the circle is 

$$A = \pi (10 - x)^2 /(2 \pi)^2$$

Area of$$A=(10 - x)^2/4\pi$$ the circle is 

then the sum of the areas is

$$A = x^2/16 + 1/2\pi \times (100 - 20x + x^2)$$ 

Next find dA/dx

$$dA/dx = x/8 + 1/(4\pi) (-20 + 2x)$$


then let the derivative = 0

$$x(1/8 + 2/4\pi) - 20/(4\pi) = 0$$

solve for x....