Question

A 6 m3 box with a perfect triangle base and lid and 3 rectangular walls is to be constructed from 2 materials.
The cost to make the base 1$/ft2 and that for the lid and walls is 4/2 $/ft2. What dimensions will result in the
most cost-effective construction?

Answer 1

The cost will be a function of the material or surface area used.
By "perfect triangle" I will assume you mean an equilateral triangle

Let each side of the base triangle be 2x, then by Pythagorus, we can easily find the height of the triangle to be √3x
and the area of the two triangles is
2(1/2)(2x)√3x) = 2√3 x^2

Let the height of the "box" be h m
the the surface area of each of the three rectangular sides is 2xh

surface area = A
= 2√3x^2 + 6xh

but √3 x^2 h = 6
h = 6/(√3x^2)

then A
= 2√3 x^2 + 6x(6/(√3x^2)
=2√3 x^2 + 36/(√3x)

Your typing of the cost of the side walls seems odd,
Why would you write it as 4/2 and not just $2 ??
I will continue as $2 per ft^2 , you will have to change the constants if otherwise

Cost = 1(2√3 x^2) + 2(36/(√3x))

dCost/dx = 4√3 x - 72/(√3 x^2) = 0 for a min of A
4√3 x = 72/(√3x^2)
12x^3 = 72
x^2 = 6
x = 6^(1/3) = 1.81712...
h = 6/(√3x^2) = 1.049115....

round to whatever accuracy you need.