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July 23rd, 2018, 10:30 PM   #1
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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 01:19 AM.
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July 23rd, 2018, 11:00 PM   #2
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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 $\displaystyle s$?
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July 24th, 2018, 01:27 AM   #3
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Which surfaces have to be painted?
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July 24th, 2018, 04:38 AM   #4
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Quote:
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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 10:31 AM.
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July 24th, 2018, 07:39 AM   #5
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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.
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July 24th, 2018, 10:44 AM   #6
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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 $\displaystyle s$?
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July 24th, 2018, 07:33 PM   #7
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Quote:
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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?
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July 24th, 2018, 08:04 PM   #8
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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.
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July 25th, 2018, 01:54 AM   #9
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Quote:
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We reject the negative value . . .
Why?
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July 25th, 2018, 06:51 PM   #10
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Why?
Length cannot be negative .
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