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June 22nd, 2015, 09:05 PM  #1 
Member Joined: Oct 2014 From: Mars Posts: 30 Thanks: 0  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 
June 22nd, 2015, 10:24 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
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. 
June 22nd, 2015, 11:16 PM  #3 
Member Joined: Oct 2014 From: Mars Posts: 30 Thanks: 0 
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? 
June 22nd, 2015, 11:29 PM  #4 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  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 02:29 AM. 
June 23rd, 2015, 02:01 AM  #5 
Member Joined: Oct 2014 From: Mars Posts: 30 Thanks: 0 
??? 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 02:04 AM. 
June 23rd, 2015, 02:05 AM  #6 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Can you let us know EXACTLY what you're trying to do?
Last edited by Azzajazz; June 23rd, 2015 at 02:16 AM. 
June 23rd, 2015, 02:33 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
conversionRatio = square root of (desiredArea/originalArea) newWidth = oldWidth * conversionRatio etc. 
June 23rd, 2015, 03:12 AM  #8 
Member Joined: Oct 2014 From: Mars Posts: 30 Thanks: 0  
June 24th, 2015, 10:18 PM  #9 
Senior Member Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 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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