optimization problem ..

[change_for_better]

Hello,

Can you help me to understand the question..Ineed clarification and hints to solve the question..

A circle and a square are to be constructed from a piece of a wire of length l.

1-give an expression for the total area of the square and circle formed.


2-find the radius of the circle and side of the square that make their areas equal.


3-find values of the radius of the circle and side of the square which give the largest and smallest total area.

4-If insted of the circle another square is formed.Find the values of the sides of the square that yeild the largest and smallest total area.

 

[change_for_better]

I solve (1) & (2) but I have difficulties on (3) &(4)

1-give an expression for the total area of the square and circle formed.

we assume that :

length of wire=circumference of circle+perimeter of square

. . . . .\(\displaystyle L\, =\, x\, +\, (L\, -\, x)\)

Area of square= a^2

*To know the length of side of square:

perimeter of square=4*a

a= perimeter/4

. . . . .\(\displaystyle a\, =\, \frac{L\, -\, x}{4}\)

Area of square

. . . . .\(\displaystyle a^2\, =\, \frac{(L\, -\, x)^2}{16}\)

Area of circle=pi*r^2

To know the radius:

r= circumference of circle / 2 PI

Area of circle = \(\displaystyle \frac{x^2}{4\pi}\)

Total Area=Area of square +Area of circle =

. . . . .\(\displaystyle \frac{(L\, -\, x)^2}{16}\, +\, \frac{x^2}{4\pi}\)

===========================================

2-find the radius of the circle and side of the square that make their areas equal.

Area of square=Area of circle :

. . . . .\(\displaystyle \frac{(L\, -\, x)^2}{16}\, +\, \frac{x^2}{4\pi}\)

The value of radius and side of square :

. . . . .\(\displaystyle a\, =\, \sqrt{\pi}r\)

. . . . .\(\displaystyle r\, =\, \frac{a}{\sqrt{\pi}}\)

___________________________
Edited by stapel -- Reason for edit: readability of formulas

 

[galactus]

The length of the wire is L.

Let x be the part of the wire used for the circle and y be the portion for the square.

Therefore, x+y=L

Let r=radius of circle and s be a side length of the square.

\(\displaystyle \text{Area}={\pi}r^{2}+s^{2}\)

where \(\displaystyle r=\frac{x}{2\pi}, \;\ s=\frac{y}{4}\)

because x is the circumference of the circle and y is the perimeter of the square.

Thus, \(\displaystyle A=\frac{x^{2}}{4\pi}+\frac{y^{2}}{16}\)

But \(\displaystyle x+y=L, \;\ y=L-x\)

\(\displaystyle A=\frac{x^{2}}{4\pi}+\frac{(L-x)^{2}}{16}\)

\(\displaystyle \frac{dA}{dx}=\frac{({\pi}+4)x-L{\pi}}{8\pi}\)

Set to 0 and solve for x gives \(\displaystyle x=\frac{L{\pi}}{{\pi}+4}\)

If \(\displaystyle x=0, \;\ \frac{L{\pi}}{{\pi}+4}, \;\ L\), then \(\displaystyle A=\frac{L^{2}}{16}, \;\ \frac{L^{2}}{4({\pi}+4)}, \;\ \frac{L^{2}}{4\pi}\)

You should find that we have a maximum when x=L (the entire wire is used for the circle).

A minimum when \(\displaystyle x=\frac{L{\pi}}{{\pi}+4}\)

If another square is formed instead of a circle, then divide the wire in half.

\(\displaystyle (\frac{x}{4})^{2}+(\frac{L-x}{4})^{2}\)

\(\displaystyle \frac{d}{dx}[(\frac{x}{4})^{2}+(\frac{L-x}{4})^{2}]=\frac{x}{4}-\frac{L}{8}\)

\(\displaystyle \frac{x}{4}-\frac{L}{8}=0\)

\(\displaystyle x=\frac{L}{2}\)

 

[change_for_better]

The length of the wire is L.

Let x be the part of the wire used for the circle and y be the portion for the square.

Therefore, x+y=L

Let r=radius of circle and s be a side length of the square.

\(\displaystyle \text{Area}={\pi}r^{2}+s^{2}\)

where \(\displaystyle r=\frac{x}{2\pi}, \;\ s=\frac{y}{4}\)

because x is the circumference of the circle and y is the perimeter of the square.

Thus, \(\displaystyle A=\frac{x^{2}}{4\pi}+\frac{y^{2}}{16}\)

But \(\displaystyle x+y=L, \;\ y=L-x\)

\(\displaystyle A=\frac{x^{2}}{4\pi}+\frac{(L-x)^{2}}{16}\)

\(\displaystyle \frac{dA}{dx}=\frac{({\pi}+4)x-L{\pi}}{8\pi}\)

Set to 0 and solve for x gives \(\displaystyle x=\frac{L{\pi}}{{\pi}+4}\)

If \(\displaystyle x=0, \;\ \frac{L{\pi}}{{\pi}+4}, \;\ L\), then \(\displaystyle A=\frac{L^{2}}{16}, \;\ \frac{L^{2}}{4({\pi}+4)}, \;\ \frac{L^{2}}{4\pi}\)

You should find that we have a maximum when x=L (the entire wire is used for the circle).

A minimum when \(\displaystyle x=\frac{L{\pi}}{{\pi}+4}\)

If another square is formed instead of a circle, then divide the wire in half.

\(\displaystyle (\frac{x}{4})^{2}+(\frac{L-x}{4})^{2}\)

\(\displaystyle \frac{d}{dx}[(\frac{x}{4})^{2}+(\frac{L-x}{4})^{2}]=\frac{x}{4}-\frac{L}{8}\)

\(\displaystyle \frac{x}{4}-\frac{L}{8}=0\)

\(\displaystyle x=\frac{L}{2}\)

Can you tell me the details How you obtain the derivative of total area in (3)because I obtain different result


and I don't understand the following

If another square is formed instead of a circle, then divide the wire in half.

\(\displaystyle (\frac{x}{4})^{2}+(\frac{L-x}{4})^{2}\)

 

[galactus]

Do you mean differentiate \(\displaystyle A=\frac{x^{2}}{4\pi}+\frac{(L-x)^{2}}{16}\)?.

Okey-doke. But that is correct.

\(\displaystyle \frac{d}{dx}[\frac{x^{2}}{4\pi}]=\frac{x}{2\pi}\)

\(\displaystyle \frac{d}{dx}[\frac{1}{16}(L-x)^{2}]=\frac{1}{16}(2)(L-x)(-1)=\frac{1}{8}(x-L)\)

Put them together:

\(\displaystyle \frac{x}{2\pi}+\frac{1}{8}(x-L)=\frac{({\pi}+4)x-L{\pi}}{8\pi}\)

There it is.

As for the last problem regarding if the other was another square instead of a circle.

Part of the wire is the same as before. It has area \(\displaystyle (\frac{x}{4})^{2}\)

The other portion of the wire is L-x. So, it's area is \(\displaystyle (\frac{L-x}{4})^{2}=\frac{(L-x)^{2}}{16}\)

\(\displaystyle \frac{d}{dx}[\frac{x^{2}}{16}+\frac{(L-x)^{2}}{16}]=\frac{x}{4}-\frac{L}{8}\)

\(\displaystyle \frac{x}{4}-\frac{L}{8}=0\)

\(\displaystyle x=\frac{L}{2}\)