Minimizing the surface area of an open box

[Specter]

Homework Statement

An open topped box with a square base has the capacity of $32m^2$. Find the dimensions that will minimize the surface area of the box.

Homework Equations

The Attempt at a Solution

I was told these are the dimensions, but I can't picture them in my head at all.
$\displaystyle A(x)=1LW+2LH+2WH$

$\displaystyle =1(x)(x)+2(x)(h)+2(x)(h)$

$\displaystyle =x^2+2xh+2xh$

$\displaystyle =x^2+4xh$

Now the formula for volume:

I'm not sure why but I need to isolate H.

$\displaystyle V=LWH$

$\displaystyle 32=(x)(x)(h)$

$\displaystyle x^2h=32$

$\displaystyle h=\frac {32} {x^2}$

Now I can substitue this into the function for the dimensions of the box

$\displaystyle A(x)=x^2+4x (\frac {32} {x^2})$

$\displaystyle =x^2+128x^{-2}$

Not sure where to go from here. Just to get to this took me forever. I can't find anything else in my lesson.

 

[scottdave]

So I'm on my phone and can't draw a pic right now. The "open box" will have 5 faces. The bottom area is Length x Width. One of the sides area is Length x Height. The opposite side has the same area, so multiply by 2. Then one adjacent side is Width x Height, and the other is the same so there is the other multiply by 2. That is the surface area (what you want to minimize.

 

[Ray Vickson]

Homework Statement

An open topped box with a square base has the capacity of $32m^2$. Find the dimensions that will minimize the surface area of the box.

Homework Equations

The Attempt at a Solution

I was told these are the dimensions, but I can't picture them in my head at all.
$\displaystyle A(x)=1LW+2LH+2WH$

$\displaystyle =1(x)(x)+2(x)(h)+2(x)(h)$

$\displaystyle =x^2+2xh+2xh$

$\displaystyle =x^2+4xh$

Now the formula for volume:

I'm not sure why but I need to isolate H.

$\displaystyle V=LWH$

$\displaystyle 32=(x)(x)(h)$

$\displaystyle x^2h=32$

$\displaystyle h=\frac {32} {x^2}$

Now I can substitue this into the function for the dimensions of the box

$\displaystyle A(x)=x^2+4x (\frac {32} {x^2})$

$\displaystyle =x^2+128x^{-2}$

Not sure where to go from here. Just to get to this took me forever. I can't find anything else in my lesson.

Why , oh why would you change from perfectly good variable names $L, W, H$ to un-informative names like $x$, etc.? You are given that $32= L H W$ and you want to minimize the area $L\,W+2\,L\, H+2\, W H$. You are also given that $W = L$, so you might as well replace $W$ by $L$ and get the problem of minimizing $L^2 + 4 L H,$ subject to the restriction $L^2 H = 32.$ That restriction implies
$$H = \frac{32}{L^2}, $$
so the area is
$$A = L^2 + 4 L \frac{32}{L^2} = L^2 + \frac{128}{L}.$$
Minimization of this $A$ is a standard calculus problem.

Note: you mistakenly had $128/L^2$ in my notation --- so check your algebra.

 

[Specter]

So I'm on my phone and can't draw a pic right now. The "open box" will have 5 faces. The bottom area is Length x Width. One of the sides area is Length x Height. The opposite side has the same area, so multiply by 2. Then one adjacent side is Width x Height, and the other is the same so there is the other multiply by 2. That is the surface area (what you want to minimize.

Yes this makes it easier for me to visualize thank you.

 

[Specter]

Why , oh why would you change from perfectly good variable names $L, W, H$ to un-informative names like $x$, etc.? You are given that $32= L H W$ and you want to minimize the area $L\,W+2\,L\, H+2\, W H$. You are also given that $W = L$, so you might as well replace $W$ by $L$ and get the problem of minimizing $L^2 + 4 L H,$ subject to the restriction $L^2 H = 32.$ That restriction implies
$$H = \frac{32}{L^2}, $$
so the area is
$$A = L^2 + 4 L \frac{32}{L^2} = L^2 + \frac{128}{L}.$$
Minimization of this $A$ is a standard calculus problem.

Note: you mistakenly had $128/L^2$ in my notation --- so check your algebra.

For the variables being replaced, I was following how it was done in my book. I understand what you posted up until you got the restriction (constraint?) $L^2H=32$. How did you find this? Edit: Nevermind that's just the formula for volume. I'm continuing to work on the quesiton.

 

[PeroK]

$\displaystyle A(x)=x^2+4x (\frac {32} {x^2})$

##=x^2+128x^{-2} ##

Not sure where to go from here. Just to get to this took me forever. I can't find anything else in my lesson.

Note that you have mistake here. The term I've underlined should be $x^{-1}$

What you need to do now is minimise $A(x)$. Any ideas?

 

[Specter]

Note that you have mistake here. The term I've underlined should be $x^{-1}$

What you need to do now is minimise $A(x)$. Any ideas?

Using the help that @Ray Vickson gave me this is what I've done:

This is probably a stupid question but...What I'm confused about is why W=L. I wasn't given the dimensions of the box so why couldn't they be different from each other. Is it just because it specifiys a square base in the question?

$\displaystyle 32m^2 = LHW$

$A(x)=1LW+2LH+2WH$

$\displaystyle L^2=4LH$

$\displaystyle V=LWH = LLH$

$\displaystyle L^H=32m^2$

Isolate H:

$\displaystyle H=\frac {32}{L^2}$

$\displaystyle A(x)=L^2+4l(\frac {32} {L})$

$\displaystyle A(x)=L^2+128L^{-1}$

Find derivative and set to zero to find L:

$\displaystyle A'(x)=2L-\frac {128} {L^2} =0$

$\displaystyle 2L=\frac {128} {L^2}$

$\displaystyle 2L^3=128$

$\displaystyle L^3=64$

$\displaystyle L=4$

Plug L into equation for H solve for H:

$\displaystyle H=\frac {32} {(4)^2}$

$\displaystyle H=2$

The dimensions of the box are 4 x 4 x 2 m.

 

[PeroK]

This is probably a stupid question but...What I'm confused about is why W=L. I wasn't given the dimensions of the box so why couldn't they be different from each other. Is it just because it specifiys a square base in the question?

Yes, that's what a square is! If $W \ne L$ then it's not a square base.

 

[Specter]

Yes, that's what a square is! If $W \ne L$ then it's not a square base.

Thank you :)

 

[StoneTemplePython]

it's worth pointing out that for surface vs volume optimization problems, you can solve/prove just about all of these with

$\text{GM} \leq \text{AM}$. For this and many of these problem Maclaurin's Inequality is a bit easier to wield, but with some insight, the more basic, and ubiquitous, $\text{GM} \leq \text{AM}$ works just fine.

Step 1: you have $v = L^2 H$ as a constraint and are minimizing $L^2 + 4LH$. The Left hand side constraint looks an elementary symmetric function $e_3$ of $\big(L, L, H\big)$ but the right hand side does not look like $e_2$ of them. (i.e. $e_2$ is product of 2 chosen at a time summing over the $\binom{n}{2}=\binom{3}{2}=3$ combinations)

Step 2: Consider a change of variables so that you have $S:= 2H$. We now have $0 \lt \frac{1}{2}v = \frac{1}{2}L^2 S = \frac{1}{2}e_3\big(L, L, S\big)$ as your constraint and $L^2 + LS +SL = e_2\big(L, L, S\big)$ on the right hand side which is what we want to minimize. The result is almost immediate with Maclaurin here.

Step 3: for the more basic approach consider squaring (which is no problem since everything is positive) and creative use of bunching:

$\big(\frac{1}{2}v\big)^2 = \frac{1}{2^2}\big(L^2 S\big)^2 = \frac{1}{2^2}\big((L^2)(LS)(SL)\big)$

Final Step: take cube roots and get

$\big(\frac{1}{2}v\big)^\frac{2}{3} = \frac{1}{2^\frac{2}{3}}\big((L^2)(LS)(SL)\big)^\frac{1}{3} \leq \frac{1}{3}\cdot \frac{1}{2^\frac{2}{3}}\big(L^2 + LS + SL\big)$

by $\text{GM} \leq \text{AM}$, with equality iff $L = S$ $\to L = 2H$. The equality conditions are key. You can also simplify with a little algebra and get the actual minimum from here.

 

[Ray Vickson]

Using the help that @Ray Vickson gave me this is what I've done:

This is probably a stupid question but...What I'm confused about is why W=L. I wasn't given the dimensions of the box so why couldn't they be different from each other. Is it just because it specifiys a square base in the question?

$\displaystyle 32m^2 = LHW$

$A(x)=1LW+2LH+2WH$

$\displaystyle L^2=4LH$

$\displaystyle V=LWH = LLH$

$\displaystyle L^H=32m^2$

Isolate H:

$\displaystyle H=\frac {32}{L^2}$

$\displaystyle A(x)=L^2+4l(\frac {32} {L})$

$\displaystyle A(x)=L^2+128L^{-1}$

Why do you write $A(x)$ when there is no variable $x$ anywhere in the formula?

BTW: instead of constantly using "displaystyle" you would get cleaner results (and a lot easier to type, too!) by using a displayed equation construction
$ $ .. your formula $ $ (with no space between the two $ at the start and the end). The intended use of "displaystyle" is (usually) within a displayed equation to over-ride the font-reducing rules that TeX sometimes uses when typesetting multi-line formulas; it prevents the fonts from becoming too small. For instance, it you entered your final three equations as a three-line array, here is what you would get
$$ \left. \begin{array}{rll}
H = & \frac{32}{L^2}, & \text{so} \\
A = & L^2 + 4 L \left( \frac{32}{L^2} \right), & \text{hence} \\
A = & L^2 + 128 L^{-1} &
\end{array} \right \} \; \Leftarrow \; \text{without "displaystyle"}
$$
$$ \left. \begin{array}{rll}
H = & \displaystyle \frac{32}{L^2}, & \text{so}\\
A = & \displaystyle L^2 + 4 L \left( \frac{32}{L^2} \right), & \text{hence} \\
A = & L^2 + 128 L^{-1} &
\end{array} \right \} \; \Leftarrow \; \text{with "displaystyle"}
$$

 

[scottdave]

They tell you it has a square base, so that is why we have L=W. Also, I just noticed in your problem statement it has a capacity (volume) of 32 m², but volume should be m³.