Maths? Help please D;

A closed box in the form of a rectangular prism is constructed such that one side of the base is twice as long as the other side. The box must have a volume of exactly 243 cm^3. Determine the dimensions of the box so that a minimum amount of material is used in its construction.



I know..this involves some sort of substitution method along with derivatives.....do i form two equations based on what the question says? HELP PLEASE !

Write out the basic equations of surface area and volume:

V = lwh
A = 2lw+2lh+2wh

Where V is the volume, l is the length, w is the width, and h is the height.

Write what you are given:

V = 243
l = 2w

Plugging these into the volume equation and solve for h:

V = lwh
(243) = (2w)wh
243 = (2w^2)h
h = 243/(2w^2)

Plug these values into the area formula:

A = 2lw + 2lh + 2wh
A = 2(2w)w + 2(2w)(243/(2w^2)) + 2w(243/(2w^2))
A = 4w^2 + 729/w

Once it has been simplified, take the derivative, set it equal to zero, and solve for w:

dA/dw = 8w - 729/w^2
0 = 8w - 729/w^2
w = 4.5

You can now plug this into the equations at the top to solve for the other dimensions.

w = 4.5 cm
l = 9.0 cm
h = 6.0 cm

Let x = box length. This makes the box's width x/2.
Volume = length * width * height
243 = x(x/2)h
h = 486/x^2
Surface area = 2 * length * width + 2 * length * height + 2 * width * height
Let f(x) = surface area.
f(x) = 2x(x/2) + 2x(486/x^2) + 2(x/2)486/x^2
f(x) = x^2 + 972/x + 486/x
f(x) = x^2 + 1458/x
Minimum occurs where f '(x) = 0 and f ''(x) > 0.
f '(x) = 2x - 1458/x^2
0 = 2x - 1458/x^2
1458/x^2 = 2x
1458 = 2x^3
729 = x^3
x = 9 or x = two imaginary numbers that can be discarded.
f ''(x) = 2 + 2919/x^3
f ''(9) = 2 + 2919/9^3 > 0 so a minimum occurs at x = 9
length = 9 cm
width = 9/2 cm
height = 486/9^2 = 486/81 = 6 cm