Find Optimal Size for Jar with Rectangular Base & Sides for Lowest Material Cost

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In summary: We can find the maximum area by finding the value of w such that $\displaystyle A = l\,w + 2l\,h + 2w\,h = V$
  • #1
Petrus
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You will produce a jar with rectangular base and rectangular sides. aLL OF IT
be manufactured in the same material. What size should the jar have on the cost of materials to be as small as possible?
The length of the can should be 1.5 times the width. The jar should hold 1 liter.
here we got a picture
2vdodbk.png
\(\displaystyle V=1000cm^3\)
\(\displaystyle h=\frac{1.5x^2}{1000cm^3}\)
Regards,
\(\displaystyle |\pi\rangle\)
 
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  • #2
Petrus said:
You will produce a jar with rectangular base and rectangular sides. aLL OF IT
be manufactured in the same material. What size should the jar have on the cost of materials to be as small as possible?
The length of the can should be 1.5 times the width. The jar should hold 1 liter.
here we got a picture
2vdodbk.png
I got \(\displaystyle A=6x^2\) and \(\displaystyle V=1000cm^3\)
\(\displaystyle h=\frac{1.5x^2}{1000cm^3}\)
Regards,
\(\displaystyle |\pi\rangle\)

First of all, am I correct in assuming that your box is to be open on the top?

Also, I'm assuming that your use of the word "can" you meant to mean the box? (Cans are usually cylindrical, not rectangular...)
 
  • #3
Prove It said:
First of all, am I correct in assuming that your box is to be open on the top?

Also, I'm assuming that your use of the word "can" you meant to mean the box? (Cans are usually cylindrical, not rectangular...)
yes the picture is open and they never mention closed, and yes box
 
  • #4
Petrus said:
You will produce a jar with rectangular base and rectangular sides. aLL OF IT
be manufactured in the same material. What size should the jar have on the cost of materials to be as small as possible?
The length of the can should be 1.5 times the width. The jar should hold 1 liter.
here we got a picture
2vdodbk.png
\(\displaystyle V=1000cm^3\)
\(\displaystyle h=\frac{1.5x^2}{1000cm^3}\)
Regards,
\(\displaystyle |\pi\rangle\)

OK, well for starters, the amount of whatever material the box is made of is the same as the surface area of the box. Since the box has an open top, that means $\displaystyle \begin{align*} A = l\,w + 2l\,h + 2w\,h \end{align*}$.

I agree, $\displaystyle \begin{align*} V = 1000\,\textrm{cm}^3 \end{align*}$, so that means $\displaystyle \begin{align*} l\,w\,h = 1000 \end{align*}$. We're also told that the length should be 1.5 times the width, so $\displaystyle \begin{align*} l = \frac{3w}{2} \end{align*}$, giving

$\displaystyle \begin{align*} l\,w\,h &= 1000 \\ \left( \frac{3w}{2} \right) \, w\, h &= 1000 \\ \frac{3w^2}{2} \, h &= 1000 \\ h &= \frac{2000}{3w^2} \end{align*}$

Substituting into the formula for the surface area we have

$\displaystyle \begin{align*} A &= l\,w + 2l\,h + 2w\,h \\ &= \left( \frac{3w}{2} \right) \, w + 2\left( \frac{3w}{2} \right) \left( \frac{2000}{3w^2} \right) + 2w \, \left( \frac{2000}{3w^2} \right) \\ &= \frac{3w^2}{2} + \frac{2000}{w} + \frac{4000}{3w} \\ &= \frac{3}{2}w^2 + \frac{10\,000}{3} w^{-1} \end{align*}$

So now how can you find the maximum area?
 
  • #5
Prove It said:
OK, well for starters, the amount of whatever material the box is made of is the same as the surface area of the box. Since the box has an open top, that means $\displaystyle \begin{align*} A = l\,w + 2l\,h + 2w\,h \end{align*}$.

I agree, $\displaystyle \begin{align*} V = 1000\,\textrm{cm}^3 \end{align*}$, so that means $\displaystyle \begin{align*} l\,w\,h = 1000 \end{align*}$. We're also told that the length should be 1.5 times the width, so $\displaystyle \begin{align*} l = \frac{3w}{2} \end{align*}$, giving

$\displaystyle \begin{align*} l\,w\,h &= 1000 \\ \left( \frac{3w}{2} \right) \, w\, h &= 1000 \\ \frac{3w^2}{2} \, h &= 1000 \\ h &= \frac{2000}{3w^2} \end{align*}$

Substituting into the formula for the surface area we have

$\displaystyle \begin{align*} A &= l\,w + 2l\,h + 2w\,h \\ &= \left( \frac{3w}{2} \right) \, w + 2\left( \frac{3w}{2} \right) \left( \frac{2000}{3w^2} \right) + 2w \, \left( \frac{2000}{3w^2} \right) \\ &= \frac{3w^2}{2} + \frac{2000}{w} + \frac{4000}{3w} \\ &= \frac{3}{2}w^2 + \frac{10\,000}{3} w^{-1} \end{align*}$

So now how can you find the maximum area?
we derivate it then equal zero, that means \(\displaystyle w=\frac{10}{3^{2/3}}\) is that correct?

Regards,
\(\displaystyle |\pi\rangle\)
 

FAQ: Find Optimal Size for Jar with Rectangular Base & Sides for Lowest Material Cost

What is the objective of finding the optimal size for a jar with a rectangular base and sides?

The objective is to determine the dimensions of the jar that will require the lowest amount of material to produce, thus reducing production costs.

How is the lowest material cost calculated for a jar with a rectangular base and sides?

The lowest material cost is calculated by considering the surface area of the jar's base and sides, as well as the thickness of the material used.

What factors should be taken into account when finding the optimal size for a jar with a rectangular base and sides?

The factors that should be considered include the desired volume of the jar, the type and cost of the material used, and any constraints on the dimensions (e.g. maximum height or width).

Are there any mathematical equations or formulas that can help determine the optimal size for a jar with a rectangular base and sides?

Yes, there are various mathematical equations and formulas that can be used, such as the surface area formula for a rectangular prism and the volume formula for a rectangular prism. Additionally, optimization techniques such as linear programming can also be applied.

How can finding the optimal size for a jar with a rectangular base and sides benefit a company or individual?

Finding the optimal size can lead to cost savings for a company or individual, as it reduces the amount of material needed for production. This can also result in a more efficient and sustainable use of resources.

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