# Percentage increase in area of rectangle given increase in perimeter

• Aug 29th 2010, 09:26 PM
darkvelvet
Percentage increase in area of rectangle given increase in perimeter
One side of a rectangle is three times the other. If the perimeter increases by 2%, what is the percentage increase in area?

I've started with these few lines:
Perimeter, P = 2x + 2x + x + x = 8x
Area, 3x^2 = 3(p^2)/64
delta P= 2%
delta A = dA/dP . delta P
delta A = 3/64 (2p). 2%

The answer is 4% but it doesn't look like delta A will be 4%!

Any clues where I went wrong? Thanks.
• Aug 29th 2010, 09:44 PM
pickslides
Quote:

Originally Posted by darkvelvet
One side of a rectangle is three times the other.

$P = 3x+3x +x+x$
• Aug 29th 2010, 09:56 PM
Educated
Quote:

Originally Posted by darkvelvet
P = 2x + 2x + x + x = 8x
Any clues where I went wrong? Thanks.

How does 2+2+1+1 = 8?
• Aug 29th 2010, 11:39 PM
darkvelvet
Sorry, it should be like this:

P = 3x + 3x + x + x = 8x
• Aug 30th 2010, 12:45 AM
Area and length of similar shapes
Hello darkvelvet
Quote:

Originally Posted by darkvelvet
One side of a rectangle is three times the other. If the perimeter increases by 2%, what is the percentage increase in area?

I've started with these few lines:
Perimeter, P = 2x + 2x + x + x = 8x
Area, 3x^2 = 3(p^2)/64
delta P= 2%
delta A = dA/dP . delta P
delta A = 3/64 (2p). 2%

The answer is 4% but it doesn't look like delta A will be 4%!

Any clues where I went wrong? Thanks.

First, a few basics:
• The percentage increase in a quantity is the actual increase divided by the original amount, multiplied by 100. So you should have $\dfrac{\delta P}{P}\times100=2$, or $\dfrac{\delta P}{P}=2$%, not $\delta P = 2$%.
.
• Since the rectangle always has the same shape, its area is proportional to the square of the length of one of its sides, and therefore is proportional to the square of its perimeter. So $A = kP^2$.

Differentiating this we get:
$\dfrac{dA}{dP}=2kP\approx\dfrac{\delta A}{\delta P}$

$\Rightarrow \dfrac{\delta A}{A} \approx\dfrac{2kP\delta P}{A}\approx \dfrac{2kP\delta P}{kP^2}}$
$\approx2\dfrac{\delta P}{P}$

So if $\dfrac{\delta A}{A}=2$%, then $\dfrac{\delta P}{P}}\approx 4$%.

In fact, it's even easier without calculus. Note that, if the lengths of corresponding sides of two similar shapes are in the ratio $1:n$, then their areas are in the ratio $1:n^2$.

A 2% increase in the length means an increase in the ratio $1:1.02$; so the area increases in the ratio $1:1.02^2 = 1 : 1.0404$, or 4.04%.

This has the added advantage of being an exact answer, whereas the calculus method gives only an approximation.

• Aug 30th 2010, 01:38 AM
darkvelvet
I get all the workings except for this:
A = kP^2.

What is k?
• Aug 30th 2010, 01:48 AM
Hello darkvelvet
Quote:

Originally Posted by darkvelvet
I get all the workings except for this:
A = kP^2.

What is k?

It is the 'constant of proportionality'. If $\displaystyle A$ is proportional to $P^2$, then we can write
$A = kP^2$, where $\displaystyle k$ is a constant.

Using your original method, you (correctly) got
$A = \dfrac{3P^2}{64}$
So
$k = \frac{3}{64}$

As you'll see from my working, $\displaystyle k$ is eliminated in the subsequent equations, so you didn't really need to evaluate it.