You are an engineer in charge of designing the dimensions of a box-like building. The base is rectangular in shape with width being twice as large as length. (Therefore so is the ceiling.) The volume is to be 1944000 m3. Local bylaws stipulate that the building must be no higher than 30 m. Suppose the walls cost twice as much per m2 as the ceiling, and suppose the floor (i.e.base) costs nothing. Find the dimensions of the building that would minimize the cost.
The answer might be easy if you told us the minimum height requirement.
Since the ceiling is cheaper, the lower the height, the cheaper it will be.
Once we know that, the squarer the building, As equal as possible in length & width, the smaller the perimeter and therefore the less m2 of walls.
Hope this helps
To minimize the cost of the building, we need to find the dimensions that minimize the surface area of the walls and ceiling.
Let's assume the length of the building is l. Based on the given information, the width of the building would be 2l, and the height should be less than or equal to 30m.
The volume of a rectangular box is given by the formula V = length × width × height. Substituting the given values, we have:
1944000 = l × 2l × h
Rearranging the equation, we get:
h = 1944000 / (2l^2)
Now, we need to find the surface area of the walls and ceiling. The formula for the surface area of a rectangular box is given by:
Surface Area = 2(length × width) + 2(length × height) + 2(width × height)
Substituting the values, we have:
Surface Area = 2(l × 2l) + 2(l × h) + 2(2l × h)
Simplifying this equation, we get:
Surface Area = 4l^2 + 2lh + 4lh
Surface Area = 4l^2 + 6lh
Now, let's calculate the cost. The cost of the walls (excluding the floor) is twice as much per square meter as the ceiling. Let's assume the cost per square meter of the ceiling is C. Therefore, the cost per square meter of the walls would be 2C.
Cost = Surface Area of Walls × Cost per Square Meter of Walls + Surface Area of Ceiling × Cost per Square Meter of Ceiling
Cost = (4l^2 + 6lh) × 2C + 2(l × 2l) × C
Cost = (8l^2 + 12lh)C + 4l^2C
Cost = (12l^2 + 12lh)C
Now, to minimize the cost, we need to find the values of l and h that minimize the cost. To do so, we can find the derivative of the cost equation with respect to l and h, set them equal to zero, and solve them simultaneously to find the critical points.
Differentiating the cost equation with respect to l, we get:
d(Cost) / dl = 24l + 12hC = 0
Differentiating the cost equation with respect to h, we get:
d(Cost) / dh = 12lC = 0
Solving these equations simultaneously, we have:
24l + 12hC = 0
12lC = 0
From the second equation, we can conclude that either l = 0 or C = 0. However, l cannot be zero as it represents the length of the building.
Therefore, C = 0 is the only possible solution, which means the cost per square meter of the ceiling should be zero to minimize the cost.
Now, let's substitute C = 0 into our cost equation:
Cost = (12l^2 + 12lh) × 0
Cost = 0
Since the cost is zero, the dimensions of the building that minimize the cost are:
Length (l) = Any positive value
Width (2l) = Twice the length
Height (h) = Determined by the volume equation
Please note that although the cost is minimized, having a zero cost for the ceiling may not be practical or feasible in real-world scenarios. In such cases, you may need to consider other constraints or factors when determining the dimensions of the building.