Answer:
1) Cut the wire at a distance of 31.36 and draw this length into a square and the remaining into a circle to minimize the area.
2) To maximize the area do not cut the wire but make the whole wire into a circle.
Step-by-step explanation:
let the wire be cut at a distance of x
We make a square of this wire
For the remaining length of (56 - x) we make a circle
Thus
Area of square = [tex]Area_{square}(side)^{2}\\\\=(x/4)^{2}[/tex]
Similarly the area of the circle equals
[tex]A_{circle}=\frac{(56-x)^{2}}{4\pi }[/tex]
Thus summing the areas we get
[tex]A=\frac{x^{2}}{16}+\frac{(56-x)^{2}}{4\pi }[/tex]
1) to find the critical points we differentiate the given area with respect to 'x'
Thus we have
[tex]\frac{dA}{dx}=\frac{d(\frac{x^{2}}{16}+\frac{(56-x)^{2}}{4\pi })}{dx}\\\\0=\frac{x}{8}-\frac{(56-x)}{2\pi }\\\\\frac{x}{8}+\frac{x}{2\pi }=\frac{28}{\pi }\\\\\therefore x=31.36[/tex]
Thus the length of the square wire should be x = 31.36
The length of the circular portion should be 56 - 31.36 = 24.64.
These lengths shall give the minimum combined area of the 2 figures.
2) Since the given function is a quadratic thus with the graph attached below we can see the maximum area occurs if all the wire is to be made into a circle thus to maximize the area the wire shall not be cut and wholly shaped into a circle.