Answer:
a.x=39.2
b.Use whole wire as a circle
Step-by-step explanation:
We are given that
Length of piece of wire=70 units
Let length of wire used to make a square =x units
Length of wire used in circle=70- x
Side of square=[tex]\frac{perimeter\;of\;square}{4}=\frac{x}{4}[/tex]
Circumference of circle=[tex]2\pi r[/tex]
[tex]70-x=2\pi r[/tex]
[tex]r=\frac{70-x}{2\pi}[/tex]
Combined area of circle and square,A=[tex](\frac{x}{4})^2+\pi(\frac{70-x}{2\pi})^2[/tex]
Using the formula
Area of circle=[tex]\pi r^2[/tex]
Area of square=[tex](side)^2[/tex]
a.[tex]A=\frac{x^2}{16}+\frac{4900+x^2-140x}{4\pi}[/tex]
Differentiate w.r.t x
[tex]\frac{dA}{dx}=\frac{x}{8}+\frac{2x-140}{4\pi}[/tex]
[tex]\frac{dA}{dx}=0[/tex]
[tex]\frac{x}{8}+\frac{2x-140}{4\pi}=0[/tex]
[tex]\frac{\pi x+4x-280}{4\pi}=0[/tex]
[tex]\pi x+4x-280=0[/tex]
[tex]x(\pi+4)=280[/tex]
[tex]x=\frac{280}{\pi+4}[/tex]
x=39.2
Again differentiate w.r.t x
[tex]\frac{d^2A}{dx^2}=\frac{1}{8}+\frac{1}{2\pi}[/tex]>0
Hence, the combined area of circle and the square is minimum at x=39.2
b.When the wire is not cut and whole wire used as a circle . Then, combined area is maximum.
