Question

If an open rectangular box is to have a fixed volume of 500m3, what relative dimensions will make the surface area minimum?

Answer 1

From prior knowledge it should be shaped like a cube to minimize SA but as for proving that

You can set up a volume equation LWH = 500

And a surface area equation

SA = 2LH + 2WH + LW (I’m interpreting “open box” as only open on one end, lmk if this is wrong)

You can use the volume equation to rewrite the SA equation in terms of one variable then take the derivative of the SA equation and set it equal to 0

Then re-start, rewriting the equation again so that the two remaining variables are rewritten in terms of one variable

Then solve for the last variable using the volume equation

Answer 2

Disclaimer: I haven’t had time to work this out fully so lmk if this method isn’t working

Answer 3

@dude would you mind checking this out when you get a chance

Answer 4

Okay going to just calculate it and see what goes

\(x^2y =500\)
\(y=\frac{500}{x^2}\)

Substitution (getting rid of y)
\(SA=x^2+4xy\color{red}{\rightarrow} x^2+4x(\frac{500}{x^2})\color{red}{\rightarrow} x^2+2000x^{-1}\)

Deriving SA
\(\frac{d}{dx}(x^2+2000x^{-1})=2x-2000x^{-2}\)


Equal to 0
\(2x-\frac{2000}{x^2}=0\)
\(2x^3-2000=0\)
\(2x^3=2000\)
\(x^3=1000\)
\(x=10\)


Checking whether its minimum
\(SA''(10)=2+\frac{4000}{x^3}\)

\(=\large 2+\frac{4000}{10^3}\)

\(SA''(10)>0\), concaves up, yay

Now solving for y
\(y=\frac{500}{x^2}\)
\(\large y=\frac{500}{10^2}\)
\(y=5\)

Now checking SA
\(10^2\cdot 5=500\)


\o/

Answer 5

Nice, I got it. Thank you both. :D
Imma split my medal into 2 XD