Optimization of a equilateral-triangular prism

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miss_b

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Hello,

I am having difficulty with this question:

A chocolate company needs to manufacture cardboard containers with the shape of equilateral-triangular prisms and that have a volume of exactly 6013mL. What are the optimum dimensions (in cm) such that a minimum amound of cardboard is used?

Answers:

Triangle edge = 28.87cm
Prism length = 16.67cm

I am guessing since \(\displaystyle v = \frac{1}{2} abh\) and \(\displaystyle sa = \left(\sqrt{3} / {4}\right)s^2 + (s+s+s)h\) (I think)

So, I want to minimise the area, so this is the equation that will be differentiated, but it has 3(?) variables, height of the rectangle, area of the triangle and width of the triangle.

As you can see, I really am confused!!

Any help would be greatly appreciated.

Thankyou
Miss_B
 


Hmmm. The given dimensions do not produce a volume of exactly 6,013 ml.

Perhaps more precision is needed?

s = 28.8658 cm

a = 16.6657 cm

I get a different function for the total suface area, in terms of s.

BTW, it's helpful if you define your variables and constants, so that other people will know what you're thinking.

s = side length of triangular ends

a = length of prism

h = height

Three congruent rectangles and two congruent triangles comprise the total surface area.

The area of each rectangle is s(a).

The area of each triangle is 0.5(s)(h).

From the Pythagorean Theorem, we know that h = sqrt(3)/2 * s.

We can express a in terms of s also, because we have the given volume.

6013 = (1/2)(s)(h)(a)

a = sqrt(3)/3 * (24052/s^2)

After substituting h and a into the expressions for the three rectangles' area and two triangles' area, followed by adding the results, I get the function for total surface area A, in terms of s:

A(s) = [sqrt(3) * (s^3 + 48104)]/(2s)

Solving A'(s) = 0 leads to s = 28.8658

 
Thank you so much for your quick response.

I follow what you are saying, and I did fail to mention that the answers are rounded to two decimal places.

I'm really not sure how to substute and rearrange the area equations to get the answers:

\(\displaystyle a(triangle) = \frac{1}{2}(s) * \left(\sqrt{3} / 2\right) (s)\) (I'm not even sure that this is right)

Do you mind spelling it out in baby talk for me?? Sorry to be a pain, but I really want to understand this.
 
The volume of the prism is \(\displaystyle 6013=\frac{1}{2}BHL\). Right?. You have that much.

B=length of triangle side, H=height of triangle, L=length of container.

But we have to shed a variable in order to solve the problem.

Since we have an equilateral, that makes it a little easier.

The height of the equilateral triangle that makes up the ends is \(\displaystyle H=\frac{\sqrt{3}}{2}B\)

So, we can sub that into the volume formula and reduce it to two variables.

\(\displaystyle 6013=\frac{1}{2}(\frac{\sqrt{3}}{2}B)L\).........[1]

Now, solve this for L and sub into the surface area formula: \(\displaystyle S=3BL+\frac{\sqrt{3}}{2}B^{2}\)........[2]

Solving [1] for L gives \(\displaystyle L=\frac{24052\sqrt{3}}{3B^{2}}\)............[3]

Sub into [2] and differentiate:

\(\displaystyle \frac{dS}{dB}=\sqrt{3}B-\frac{24052\sqrt{3}}{B^{2}}\)

Set to 0 and solve for B:

\(\displaystyle \sqrt{3}B-\frac{24052\sqrt{3}}{B^{2}}=0\)

\(\displaystyle B=6013^{\frac{1}{3}}\cdot 2^{\frac{2}{3}}\approx 28.865\)

Sub this back into [3] to find the length is \(\displaystyle L=\frac{6013^{\frac{1}{3}}\sqrt{3}\cdot 2^{\frac{2}{3}}}{3}\approx 16.666\)
 
That is magic! Thank you - both of you for your time (and patience hehe).

But I do have one question...(sorry)

When substituting H into the Volume equation and solving for L - why and how does the other 'B' get eliminated?

I got my last question (optimization of a can) straight away and this one is just horrible! I think its the sqrt that is confusing me :?
 
\(\displaystyle Only \ two \ varables \ are \ needed \ since \ we \ are \ dealing \ with \ an \ equilateral \ triangle.\)

\(\displaystyle Furthermore, \ 1 \ ml \ = \ 1 \ cm^{3} \ and \ Area_{equilateral \ triangle} \ = \ \frac{\sqrt 3s^{2}}{4}, \ s \ being \ one \ of \ its \ sides.\)

\(\displaystyle Hence, \ V_{prism} \ = \ 6013 \ = \ \bigg[\frac{\sqrt 3s^{2}}{4}L\bigg], \ L \ = \ length \ of \ prism.\)

\(\displaystyle Surface \ Area \ of \ prism \ (S.A.) \ = \ (3)(s)(L) \ + \ 2\bigg[\frac{\sqrt 3s^{2}}{4}\bigg],\)

\(\displaystyle Ergo, \ L \ = \ \frac{24,052}{\sqrt 3s^{2}}, \ \implies \ S.A._{prism} \ = \ \frac{\sqrt 3(24,052)}{ s} \ + \ \frac{\sqrt 3s^{2}}{2}.\)

\(\displaystyle \frac{d[S.A.]}{ds} \ = \ -\frac{\sqrt 3(24.052)}{s^{2}} \ + \ \sqrt 3s, \ setting \ the \ slope \ = \ to \ 0, \ we \ get \ s \ = \ 28.87.\)

\(\displaystyle Therefore \ L \ = \ \frac{24.052}{\sqrt 3(28.87)^{2}} \ = \ 16.66.\)

\(\displaystyle Check: \ V_{prism} \ = \ \frac{\sqrt 3(28.87)^{2}(16.66)}{4} \ = \ 6012.695 \ - \ good \ enough \ for \ government \ work.\)

See below for graph of the Surface Area, and note its minimum occurs when s = 28.87.

[attachment=0:1vgnbhh0]zzz.jpg[/attachment:1vgnbhh0]
 

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miss_b said:
Hello,

I am having difficulty with this question:

A chocolate company needs to manufacture cardboard containers with the shape of equilateral-triangular prisms and that have a volume of exactly 6013mL. What are the optimum dimensions (in cm) such that a minimum amound of cardboard is used?

Answers:

Triangle edge = 28.87cm
Prism length = 16.67cm

I am guessing since \(\displaystyle v = \frac{1}{2} abh\) and \(\displaystyle sa = \left(\sqrt{3} / {4}\right)s^2 + (s+s+s)h\) (I think)

So, I want to minimise the area, so this is the equation that will be differentiated, but it has 3(?) variables, height of the rectangle, area of the triangle and width of the triangle.

As you can see, I really am confused!!

Any help would be greatly appreciated.

Thankyou
Miss_B
Click to expand...

Yes, the volume of the prism is

\(\displaystyle v = \frac{1}{2} abh\)

BUT....this is an equilateral triangle, so \(\displaystyle \frac {1}{2}ab\) can be expressed as \(\displaystyle \left(\sqrt{3} / {4}\right)s^2\)


Therefore, \(\displaystyle v = \left(\sqrt{3} / {4}\right)s^2 * h\)

And you know the volume must be 6103 mL.

\(\displaystyle 6103 =\left(\sqrt{3} / {4}\right)s^2 * h\)

You should be able to solve this for h in terms of s.

Then substitute the expression you get for "h" in the formula for SA....and by the way, you need 2 bases in a prism, so you'll have to multiply the area of the triangular base by 2 in the formula for SA.

You'll end up with a SA formula that involves only ONE variable....s. Then, differentiate and minimize the surface area.
 
miss_b said:
… \(\displaystyle [area](triangle) = \frac{1}{2}(s) * \left(\sqrt{3} / 2\right) (s)\) (I'm not even sure that this is right) …
Click to expand...


Yes, it's right.

If you forget the sqrt(3)/2 factor in the height of an equilateral triangle, then here's how you can reason out the height using the Pythagorean Theorem.

The height "cuts" the equilateral triangle in half.

Code: 
     .
     |\
     | \
  h  |  \   s
     |   \
     |    \
     |_____\

     (1/2)s

s^2 = [(1/2)s]^2 + h^2

Solve for h (ignore negative square roots).

h = sqrt(3)/2 * s

8-)
 
Ok...I apologize for any confusion....but I assumed that "h" was the height of the prism...not anything to do with the equilateral triangle which is the base of the prism.

Area of the equilateral triangle base = sqrt(3) * (s[sup:3mth028k]2[/sup:3mth028k] / 4)

Volume of prism = (area of base)*height

Volume of prism = [sqrt(3) * (s[sup:3mth028k]2[/sup:3mth028k] / 4) * h

You know the volume of the prism is 6013...
 
Thank you all so much. I *almost* understand it now. I have some more of the same problem to work with, so i'm sure I will have a thorough understanding soon :)

Thank you again
 
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