# Math Help - Perimeter word problem

1. ## Perimeter word problem

Posted: Sun Nov 19, 2006 11:56 pm Post subject: Perimeter word problem

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Hi,

Im having some problems with a word problem given by my instructor.

The quesiton is -

You are given 50m of rope, and cut it into two sections (x and y). With one part of the rope, you make a circle. With the other half, you form a square.
a) Create an equation to illustrate this idea and
b) maximize the area of both objects

Im having problems creating an equation for this problem. I know that

x+y=50
Peremeter of a circle - 2(Pi)r
peremeter of a square - 4L

Can anyone help me formulate the equation so I can maximize the peremeter?

Thanks!

Matt

2. Originally Posted by matt944
Posted: Sun Nov 19, 2006 11:56 pm Post subject: Perimeter word problem

--------------------------------------------------------------------------------

Hi,

Im having some problems with a word problem given by my instructor.

The quesiton is -

You are given 50m of rope, and cut it into two sections (x and y). With one part of the rope, you make a circle. With the other half, you form a square.
a) Create an equation to illustrate this idea and
b) maximize the area of both objects

Im having problems creating an equation for this problem. I know that

x+y=50
Peremeter of a circle - 2(Pi)r
peremeter of a square - 4L

Can anyone help me formulate the equation so I can maximize the peremeter?

Thanks!

Matt
the perimeter of the circle is x
The perimeter of the square is y
so $P_x+P_y=50$

Now the trick is finding a relationship between perimeter and area, let's start with the circle.

We have: $P_x=2\pi r$

Divide both sides by 2: $\frac{P_x}{2}=\pi r$

multiply both sides by r: $\frac{P_x}{2}\cdot r=\pi r^2$

Substitute to get: $\frac{P_x}{2}\cdot\frac{P_x}{2\pi}=A_x$

Thus: $\frac{P_x^2}{4\pi}=A_x$

Now let's find the area of the square, we know that: $P_y=4s$ where $s$ means each side.

Divide both sides by 4: $\frac{P_y}{4}=s$

Square both sides: $\frac{P_y^2}{16}=s^2$

Substitute: $\frac{P_y^2}{16}=A_y$

Now go back to the equation: $x+y=50$

Substitute: $P_x+P_y=50$

Thus: $P_y=50-P_x$

You are trying to maximize: $A_x+A_y$

Substitute: $\frac{P_x^2}{4\pi}+\frac{P_y^2}{16}$

Thus: $\frac{4P_x^2}{16\pi}+\frac{P_y^2\pi}{16\pi}$

Add: $\frac{4P_x^2+P_y^2\pi}{16\pi}$

Substitute: $\frac{4P_x^2+(50-P_x)^2\pi}{16\pi}$

Thus: $\frac{4P_x^2+\left(2500-100P_x+P_x^2\right)\pi}{16\pi}$

Then: $\frac{4P_x^2+2500\pi-100P_x\pi+P_x^2\pi}{16\pi}$

Group like terms: $\frac{4P_x^2+P_x^2\pi-100P_x\pi+2500\pi}{16\pi}$

Add: $\frac{(4+\pi)P_x^2-100P_x\pi+2500\pi}{16\pi}$

Now maximize.