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

[Petrus]

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

$$V=1000cm^3$$
$$h=\frac{1.5x^2}{1000cm^3}$$
Regards,
$$|\pi\rangle$$

 

[Prove It]

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

I got $$A=6x^2$$ and $$V=1000cm^3$$
$$h=\frac{1.5x^2}{1000cm^3}$$
Regards,
$$|\pi\rangle$$

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...)

 

[Petrus]

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

 

[Prove It]

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

$$V=1000cm^3$$
$$h=\frac{1.5x^2}{1000cm^3}$$
Regards,
$$|\pi\rangle$$

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?

 

[Petrus]

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 $$w=\frac{10}{3^{2/3}}$$ is that correct?

Regards,
$$|\pi\rangle$$