What are the dimensions of a box that would hold 250 cubic centimeters of juice and have a minimum surface area

The dimensions of a box that have the minium surface area for a given Volume is such that it is a cube. This is the three dimensions are equal:

V = x*y*z , x=y=z => V = x^3, that will let you solve for x,

x = ∛(V) = ∛(250cm^3) = 6.30 cm.

6.30 cm * 6.30cm * 6.30cm. This is a cube of side 6.30cm.

The demonstration of that the shape the minimize the volume of a box is cubic (all the dimensions equal) corresponds to a higher level (multivariable calculus).

I guess it is not the intention of the problem that you prove or even know how to prove it (unless you are taking an advanced course).

Nevertheless, the way to do it is starting by stating the equations for surface and apply two variable derivation to optimize (minimize) the surface.

You do not need to follow with next part if you do not need to understand how to show that the cube is the shape that minimize the surface.

If you call x, y, z the three dimensions, the surface is:

S = 2xy + 2xz + 2yz (two faces xy, two faces xz and two faces yz).

Now use the Volumen formula to eliminate one variable, let's say z:

V = x*y*z => z = V /(x*y)

=> S = 2xy + 2x [V/(xy)[ + 2y[V/(xy)] = 2xy + 2V/y + 2V/x

Now find dS, which needs the use of partial derivatives. It drives to:

dS = [2y  - 2V/(x^2)] dx + [2x - 2V/(y^2) ] dy = 0

By the properties of the total diferentiation you have that:

2y - 2V/(x^2) = 0 and 2x - 2V/(y^2) = 0

2y - 2V/(x^2) = 0 => V = y*x^2

2x - 2V/(y^2) = 0 => V = x*y^2

=> y*x^2 = x*y^2 => y*x^2 - x*y^2 = xy (x - y) = 0 => x = y

Now that you have shown that x = y.

You can rewrite the equation for S and derive it again:

S = 2xy + 2V/y + 2V/x, x = y => S = 2x^2 + 2V/x + 2V/x = 2x^2 + 4V/x

Now find S'

S' = 4x - 4V/(x^2) = 0 => V/(x^2) = x => V =x^3.

Which is the proof that the box is cubic.