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 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: 12,211 Thanks: 520 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 $\displaystyle s$? Thanks from greg1313 and topsquark July 24th, 2018, 12:27 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,375 Thanks: 2010 Which surfaces have to be painted? July 24th, 2018, 03:38 AM   #4
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 Originally Posted by MarkFL 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 $\displaystyle s$?
A = 2s^2 +4s(s+1.4) is that right?

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: 3,264 Thanks: 902 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. Thanks from greg1313 and topsquark July 24th, 2018, 09:44 AM   #6
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 Originally Posted by Country Boy 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.
We need to distribute the 4 to the 1.4 so that our quadratic is:

$\displaystyle 6s^2+5.6s-50=0$

or:

$\displaystyle 15s^2+14s-125=0$

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
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 Originally Posted by MarkFL We need to distribute the 4 to the 1.4 so that our quadratic is: $\displaystyle 6s^2+5.6s-50=0$ or: $\displaystyle 15s^2+14s-125=0$ Now we can apply the quadratic formula or complete the square. Are there any constraints on $\displaystyle s$?
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
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 Originally Posted by speedzone 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?
You are correct in that we discard the negative root. Next we consider that the length and breadth of the rectangular faces are interchanged. That is:

$\displaystyle 2s^2+4s(s-1.4)=50$

This will also give you a positive and negative root. July 25th, 2018, 12:54 AM   #9
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 Originally Posted by speedzone We reject the negative value . . .
Why? July 25th, 2018, 05:51 PM   #10
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 Originally Posted by skipjack Why?
Length cannot be negative .  Tags optimization, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post timeone Linear Algebra 1 April 24th, 2014 10:18 PM TJ44 Calculus 1 April 14th, 2012 01:14 PM Chee Calculus 2 April 9th, 2012 07:13 PM supernerd707 Calculus 1 June 13th, 2011 04:43 AM bebejay5 Calculus 2 December 13th, 2009 01:38 PM

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