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?

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.

To determine the most cost-effective construction, we need to find the dimensions that minimize the cost of materials. Let's start by analyzing the cost of the box constructed from two different materials.

Let's assume the dimensions of the triangle base are:
- Base length = b
- Height of the triangle = h

The formula to calculate the area of a triangle is:
Area = (1/2) * base * height

The cost of the triangle base would be:
Cost of triangle base = Area * Cost per unit area = (1/2) * b * h * 1

Next, let's assume the dimensions of the rectangular walls are:
- Length of each wall = l
- Width of each wall = w

The surface area of each rectangular wall would be:
Area of rectangle = length * width

The cost of each rectangular wall would be:
Cost of each rectangular wall = Area of rectangle * Cost per unit area = l * w * (4/2)

Since there are three rectangular walls, the total cost of the rectangular walls would be:
Total cost of rectangular walls = 3 * (Cost of each rectangular wall)

Finally, the cost of the lid would be the same as the cost of the rectangular walls.

Now, the volume of the box is given as 6 m3. Therefore, the volume can also be expressed using the dimensions we have assumed:
Volume of box = Area of triangle base * height of the box = (1/2) * b * h * h

Now, let's substitute the volume equation into the equation for the area of the triangle base:
6 = (1/2) * b * h * h

We can rewrite this equation as:
h^2 = 12 / (b * h)

Simplifying further:
h^3 = 12 / b

Now, let's find the cost equation by substituting the equations for the cost of the base and the cost of the rectangular walls:
Cost = (1/2) * b * h * 1 + 4/2 * (l * w) * 3 + 4/2 * (l * w)

Simplifying further:
Cost = (1/2) * b * h + 4 * (l * w) + 4 * (l * w)

Therefore, the cost equation is:
Cost = (1/2) * b * h + 8 * (l * w)

Now, to find the dimensions that result in the most cost-effective construction, we need to minimize the cost equation. We can do this by taking derivatives with respect to b, l, and w, setting them equal to zero, and solving for the dimensions.

However, since we don't have specific values for the cost per unit area or any constraints on the dimensions, we cannot provide the exact dimensions that would result in the most cost-effective construction.