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June 22nd, 2015, 08:05 PM   #1
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Question Calculate the length and width of a rectangle

I have a problem I'm trying to solve. At the moment I'm solving it using brute force, but am wondering if there is a simple formula to do it?

Imagine you have a rectangle of size 20 units X 10 units. The area (20*10) is 200 units.

Now, lets say you want to keep the width/height ratio of the rectangle the same, but you want the new area to be no more than 150 units.

How do you calculate the new width and height?

Thanks in advance
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June 22nd, 2015, 09:24 PM   #2
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In your current rectangle, the width to height ratio is 2:1. Thus, if $w$ is the width of the rectangle and $h$ is the height, we have
$\quad w = 2t$ and
$\quad h = t$
for some number $t$.

Now, we want the area to be less than or equal to 150, or
$\quad wh\leq150$

$\quad 2t^2\leq150$

$\quad t^2\leq75$

$\quad t\leq\sqrt{75} = 5\sqrt{3}$ since $w$ and $h$ are both positive.

Thus, any value $w$ and $h$ satisfying $w\leq10\sqrt{3}$ and $h\leq5\sqrt{3}$ will work.
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June 22nd, 2015, 10:16 PM   #3
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Sorry but I don't understand...

I'm not really good at math..

Could you write the formula in such a way that I could just type it into excel?
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June 22nd, 2015, 10:29 PM   #4
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Let's say you want to keep

$\displaystyle \color{blue}{w\in\mathbb{N},\ \ h\in\mathbb{N}.
\\\;\\
\dfrac{w}{h}=\dfrac{20}{10}\Longrightarrow \dfrac{w}{h}=2\Longrightarrow w\ =\ 2h \ \ \ (1)
\\\;\\
\mathcal{A}=w\cdot h \stackrel{(1)}{\Longrightarrow} \mathcal{A}=2h\cdot h\Longrightarrow \mathcal{A}=2h^2\ \ \ (2)
\\\;\\
\mathcal{A} \ < \ 150 \ \ \ (3)
\\\;\\
(2), \ (3) \Longrightarrow 2h^2<150 \Longrightarrow h^2<75 \Longrightarrow h \in \{1, \ 2,\ 3,\ 4, \ 5,\ 6, \ 7, \ 8 \} \ \ \ (4)
\\\;\\
(1),\ (4) \Longrightarrow w \in \{2, \ 4,\ 6,\ 8, \ 10,\ 12, \ 14, \ 16 \} \ \ \ (5)
\\\;\\
(4),\ (5) \Longrightarrow (w,\ h) \in \{(2,\ 1), \ (4,\ 2), \ (6,\ 3),\ ( 8,\ 4), \ (10,\ 5),\ (12, \ 6),\ (14, \ 7),\ (16, \ 8 ) \}}$

Last edited by skipjack; June 23rd, 2015 at 01:29 AM.
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June 23rd, 2015, 01:01 AM   #5
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???

I am completely lost. I'm sorry but I cannot read that.

Maybe I shouldn't have used the word 'formula'. I think what I meant was I needed an algorithm to solve the problem. So I can just feed it the width, height and desired area and have it spit out the new width and height.

Something like...

newWidth = (oldWidth*oldHeight) / desiredArea etc...

Last edited by Apple30; June 23rd, 2015 at 01:04 AM.
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June 23rd, 2015, 01:05 AM   #6
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Can you let us know EXACTLY what you're trying to do?

Last edited by Azzajazz; June 23rd, 2015 at 01:16 AM.
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June 23rd, 2015, 01:33 AM   #7
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conversionRatio = square root of (desiredArea/originalArea)
newWidth = oldWidth * conversionRatio
etc.
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June 23rd, 2015, 02:12 AM   #8
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Quote:
Originally Posted by skipjack View Post
conversionRatio = square root of (desiredArea/originalArea)
newWidth = oldWidth * conversionRatio
etc.
Thankyou!!!
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June 24th, 2015, 09:18 PM   #9
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Math Focus: Calculus
This question will bring out an inequality which will have a set of values as solution.no single solution is possible.
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