July 23rd, 2018, 09:30 PM  #1 
Newbie Joined: Mar 2018 From: Malaysia Posts: 5 Thanks: 0  Not sure if it's an Optimization question
Joe wants to construct a solid rectangular cuboid with 2 square ends. He has enough paint for a surface area of 50m^2. For artistic purposes, the length of the rectangular faces must be 1.4m longer than its breadth. If Joe wants to use up all of his paint, find the dimensions of the two possible cuboids that he can construct. How do I go about doing this? Please help! Last edited by skipjack; July 24th, 2018 at 12:19 AM. 
July 23rd, 2018, 10:00 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs 
I would let $\displaystyle s$ be the length of the sides of the square ends. Can you write the surface area of the cuboid as a function of $\displaystyle s$?

July 24th, 2018, 12:27 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,285 Thanks: 1681 
Which surfaces have to be painted?

July 24th, 2018, 03:38 AM  #4  
Newbie Joined: Mar 2018 From: Malaysia Posts: 5 Thanks: 0  Quote:
And then what do I do? Last edited by skipjack; July 24th, 2018 at 09:31 AM.  
July 24th, 2018, 06:39 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 893 
Yes, taking the length of one side of one of the "square ends" to be "s", the area of both square ends is $\displaystyle 2s^2$. Then the length of any one of the other 4 sides is s+ 1.4 so the area of such a face is $\displaystyle s(s+ 1.4)= s^2+ 1.4s$. There are 4 such faces so the area of all 4 is $\displaystyle 4s(s+ 1.4)$. The overall surface area is $\displaystyle 2s^2+ 4s(s+ 1.4)= 2s^2+ 4s^2+ 1.4s= 6s^2+ 1.4s$. Assuming you have enough paint to paint 50 square meters, then you could paint such an object for s up to $\displaystyle 6s^2+ 1.4s= 50$. Solve that quadratic equation for s, in meters. 
July 24th, 2018, 09:44 AM  #6  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs  Quote:
$\displaystyle 6s^2+5.6s50=0$ or: $\displaystyle 15s^2+14s125=0$ Now we can apply the quadratic formula or complete the square. Are there any constraints on $\displaystyle s$?  
July 24th, 2018, 06:33 PM  #7 
Newbie Joined: Mar 2018 From: Malaysia Posts: 5 Thanks: 0  So we solve the equation and we get a positive and negative value for s. We reject the negative value but question asked for dimensions of TWO possible cuboids? How is that possible?

July 24th, 2018, 07:04 PM  #8  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs  Quote:
$\displaystyle 2s^2+4s(s1.4)=50$ This will also give you a positive and negative root.  
July 25th, 2018, 12:54 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 19,285 Thanks: 1681  
July 25th, 2018, 05:51 PM  #10 
Newbie Joined: Mar 2018 From: Malaysia Posts: 5 Thanks: 0  

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