Optimation of a box

In an open box with quadratic bottom and volume V the side is called c and the hight is called h.

a) Find h expressed by x and V
Easy: V=x^2*h
so h=V/x^2

b)Express the dimentions (in terms of V) of the box so the surface is minimal.
So the surface is given by A=x^2+4xh (since open)
I could replace h from before so A=x^2+4x*(V/x^2) witch is A=x^2+4V/x but I still have a function of two variables.
When I get a funktion of one variable I only need to find the derivative and solve the equation A'(v) or A'(x) equal to 0, but my problem is the two variables.

Acording to the book the results are: x equals the third root of 2v and h=1/2x.

I hope you can help me.
Kind regards
J.

Hello, how to teach!

An open-box with a square bottom has volume \(\displaystyle V\), side \(\displaystyle x\) and height \(\displaystyle h.\)

a) Find \(\displaystyle h\) in terms of \(\displaystyle x\) and \(\displaystyle V.\)

. . Easy: .\(\displaystyle V\:=\:x^2h \quad\Rightarrow\quad h\:=\:\frac{V}{x^2}\)

b) Express the dimensions of the box (in terms of \(\displaystyle V\)) so the surface is minimal.

So the surface is given by: .\(\displaystyle A\:=\:x^2+4xh\) (since open)

I could replace \(\displaystyle h\) so \(\displaystyle A\:=\:x^2+4x\left(\frac{V}{x^2}\right) \qud\Rightarrow\quad A\:=\:x^2+\frac{4V}{x}\)
. . but I still have a function of two variables. . . . . no

When I get a function of one variable, I need to find the derivative
. . and solve the equation \(\displaystyle A'(V)= 0\,\text{ or }\,A'(x) = 0\)
But my problem is the two variables.

According to the book, the results are: .\(\displaystyle x \:=\:\sqrt[3]{2V}\,\text{ and }\,h\:=\:\frac{1}{2}x\)

Click to expand...

To minimize the surface area of the box, the volume must be given.
. . So \(\displaystyle V\) is a constant.

Then
\(\displaystyle A'(x)=2x-\frac{4v}{x^2} \)
And if we set
\(\displaystyle A'(x)=0\)
We get
\(\displaystyle 2x-\frac{4v}{x^2}=0 \implies 4v=2x^3 \implies v=\sqrt[3]{4v}\)
and the other can be found by;
\(\displaystyle 4v=2x^3 \implies v=\frac{1}{2}x^3\)
so
\(\displaystyle v(x)=x^2\cdot h \implies h=\frac{1}{2}x\)
Perfect. You saved my day!

And I had not seen the tex-bottom...