Mathematics: A piece of wire of length 7070 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to (a) minimize and (b) maximize the combined area of the circle and the square?

the answer is 16a2-4a+4a-1it can be simplified to 16a2-1, which can be factored to (4a+1)(4a-1)...

A piece of wire of length 7070 is cut, and the

DavidSuperior3475

29.01.2022

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= $\frac{perimeter\;of\;square}{4}=\frac{x}{4}$

Circumference of circle= $2\pi r$

$70-x=2\pi r$

$r=\frac{70-x}{2\pi}$

Combined area of circle and square,A= $(\frac{x}{4})^2+\pi(\frac{70-x}{2\pi})^2$

Using the formula

Area of circle= $\pi r^2$

Area of square= (side)^2

a. $A=\frac{x^2}{16}+\frac{4900+x^2-140x}{4\pi}$

Differentiate w.r.t x

$\frac{dA}{dx}=\frac{x}{8}+\frac{2x-140}{4\pi}$

$\frac{dA}{dx}=0$

$\frac{x}{8}+\frac{2x-140}{4\pi}=0$

$\frac{\pi x+4x-280}{4\pi}=0$

$\pi x+4x-280=0$

$x(\pi+4)=280$

$x=\frac{280}{\pi+4}$

x=39.2

Again differentiate w.r.t x

$\frac{d^2A}{dx^2}=\frac{1}{8}+\frac{1}{2\pi}$ >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.

0,0(0 оценок)