An architect is designing a rectangular building in which the front wall costs twice as much per linear

metre as the other 3 walls. The building is to cover 1350 m2. What dimensions must it have so that the
costs of the walls is a minimum?

write a formula using derivatives

To find the dimensions that minimize the cost of the walls, we need to define the cost function. Let's denote the length of the front wall as x and the length of the side walls as y.

Since the front wall costs twice as much per linear meter as the other three walls, the cost function can be expressed as:

C(x, y) = 2x + 3y

To eliminate one variable, we can use the given area of the building. The formula for the area of a rectangle is:

A = x * y

Substituting the value of y from the area equation into the cost function, we get:

C(x) = 2x + 3(A/x)

Now, we need to find when the cost is at a minimum by differentiating C(x) with respect to x, setting it equal to zero, and solving for x:

dC/dx = 2 - 3(A/x^2) = 0

Simplifying further:

2 = 3(A/x^2)

x^2 = 3A/2

x = sqrt(3A/2)

We can substitute this value of x back into the area equation to find the corresponding value of y:

A = x * y

1350 = sqrt(3A/2) * y

y = 1350 / sqrt(3A/2)

Thus, the dimensions that minimize the cost of the walls while covering an area of 1350 m^2 are:

x = sqrt(3A/2)
y = 1350 / sqrt(3A/2)

The given information says nothing about the height or width of the building. Without that, we do not know enough to work the problem.