# Thread: Determining the minimum and maximum (optimization problem)

1. ## Determining the minimum and maximum (optimization problem)

A piece of wire 40cm long is to be cut in two. One piece is bent to form a square and the other is bent to form a circle.

a) Determine the length of each piece of wire so t he sum of the areas is a minimum.

b) Determine the length of each piece of wire so the sum of the areas is a maximum.

How would I go about solving this?

2. Let $\displaystyle x$ be length of the wire for the Circle $\displaystyle C$, then $\displaystyle 40-x$ is the length of the wire for the square $\displaystyle S$.

observe that: $\displaystyle 2\pi r = x \rightarrow r = \frac{x}{2\pi}$. Hence the area of $\displaystyle C$ is given by $\displaystyle \pi r^2=\frac{x^2}{4\pi}$. And the area of $\displaystyle S$ is given by $\displaystyle \frac{(40-x)^2}{4^2}$ (why?)

Hence for (a)/(b) we must find $\displaystyle 0\leq x\leq 40$ such that

$\displaystyle f(x)=\frac{x^2}{4\pi} +\frac{(40-x)^2}{4^2}$ is maximal/minimal

You can see by plotting $\displaystyle f(x)$ for example that the maximum is reached by $\displaystyle x=40$, and the minimum is when $\displaystyle f'(x)=0$. (you can easily solve that)

3. Originally Posted by Dinkydoe
Let $\displaystyle x$ be length of the wire for the Circle $\displaystyle C$, then $\displaystyle 40-x$ is the length of the wire for the square $\displaystyle S$.

observe that: $\displaystyle 2\pi r = x \rightarrow r = \frac{x}{2\pi}$. Hence the area of $\displaystyle C$ is given by $\displaystyle \pi r^2=\frac{x^2}{4\pi}$. And the area of $\displaystyle S$ is given by $\displaystyle \frac{(40-x)^2}{4^2}$ (why?)

Hence for (a)/(b) we must find $\displaystyle 0\leq x\leq 40$ such that

$\displaystyle f(x)=\frac{x^2}{4\pi} +\frac{(40-x)^2}{4^2}$ is maximal/minimal

You can see by plotting $\displaystyle f(x)$ for example that the maximum is reached by $\displaystyle x=40$, and the minimum is when $\displaystyle f'(x)=0$. (you can easily solve that)
What exactly is 2pir = x --> r = x/2pi coming from? I don't really get what you mean by that I understand that the area of a circle, C, is given by pir^2, but why are you changing it to x^2/4pi (and how did you change it to that?) I also don't understand why the area of the square, S, is given by (40-x)^2 / 4^2 (I don't get where that comes from.) Sorry

4. Originally Posted by kmjt
What exactly is 2pir = x --> r = x/2pi coming from? I don't really get what you mean by that I understand that the area of a circle, C, is given by pir^2, but why are you changing it to x^2/4pi (and how did you change it to that?) I also don't understand why the area of the square, S, is given by (40-x)^2 / 4^2 (I don't get where that comes from.) Sorry
an extended explanation of what Dinkydoe correctly posted ...

the wire of length 40 cm is cut into two pieces.

one piece has length $\displaystyle x$

the other piece has length $\displaystyle 40-x$

the piece of length $\displaystyle x$ is bent to form a circle ... this tells you that $\displaystyle x$ = the circumference of the circle, or $\displaystyle x = 2\pi r$

solve for $\displaystyle r$ in terms of $\displaystyle x$ ...

$\displaystyle r = \frac{x}{2\pi}$

area of the circle is $\displaystyle A = \pi r^2$

substitute $\displaystyle \frac{x}{2\pi}$ for $\displaystyle r$ in the circle area equation ...

$\displaystyle A = \pi \cdot \left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi}$

... the second piece has length of $\displaystyle 40-x$ which is bent to form a square.

since $\displaystyle 40-x$ = perimeter of the square, each side of the square is $\displaystyle s = \frac{40-x}{4}$

area of the square is $\displaystyle s^2 = \frac{(40-x)^2}{16}$

total area = circle area + square area

so, total area of both the circle and the square in terms of $\displaystyle x$ is ...

$\displaystyle A = \frac{x^2}{4\pi} + \frac{(40-x)^2}{16}$

find $\displaystyle \frac{dA}{dx}$ and determine the value(s) of x that optimizes the combined area.

5. Why aren't you simplifying

into (1600-x^2) / 16?

6. Originally Posted by kmjt
Why aren't you simplifying

into (1600-x^2) / 16?
You may want to revisit how to expand polynomials. $\displaystyle (40 - x)^2$ is not equal to $\displaystyle 1600 - x^2$.

7. Originally Posted by kmjt
Why aren't you simplifying

into (1600-x^2) / 16?
is this what you meant?

$\displaystyle \frac{(40-x)^2}{16} = \frac{1600-80x+x^2}{16}$

note that expansion of a binomial is not "simplifying" it.

besides, taking the derivative of $\displaystyle \frac{(40-x)^2}{16}$ is rather simple ...

$\displaystyle \frac{d}{dx} \frac{(40-x)^2}{16} = \frac{x-40}{8}$

8. Still having a bit of trouble with this. This is what I have done so far:

x represents the circumference of the circle
40-x represents the perimeter of the square

the circumference of the circle = 2pir
x=2pir

r= x / 2pi

Area of a circle = pir^2

subbing in r:
A = pi (x/2pi)^2

Area of circle = x^2 / 4pi

the length of one side of the square = (40-x) / 4

Area of a square = lw
= [(40-x)/4)][(40-x)/4)]
=(40-x)(40-x) / 16
= x^2 -80x +1600 / 16

So the area of the circle and square = x^2 / (4pi) + (x^2 -80x +1600) / 16

I then derived that to:

A' = x / (2pi) + (32x - 1280) / 256

And I know to find the max and min I have to first find the critical numbers and than make a table using intervals of my critical numbers. But how would I set x / (2pi) + (32x - 1280) / 256 equal to 0? And what would my intervals be in my table?

9. I told you it was easier if you left the area function in the following form ...

$\displaystyle A = \frac{x^2}{4\pi} + \frac{(40-x)^2}{16}$

$\displaystyle \frac{dA}{dx} = \frac{x}{2\pi} - \frac{(40-x)}{8} = 0$

$\displaystyle \frac{x}{2\pi} = \frac{(40-x)}{8}$

$\displaystyle 8x = 80\pi - 2\pi x$

$\displaystyle 4x = 40\pi - \pi x$

$\displaystyle 4x + \pi x = 40\pi$

$\displaystyle x(4 + \pi) = 40\pi$

$\displaystyle x = \frac{40\pi}{4+\pi} \approx 17.6$

there is your single critical value ... you should be able to tell this value yields a minimum, based on the fact the original area function is a quadratic.

it should also tell you that the area function has an endpoint maximum.