A box of constant volume c is to be twice long as it is wide . the material on the top and four sides cost three times as much per square metre as that in the bottom. what are most economic dimensions ?
To find the most economic dimensions of the box, we need to consider the cost of the materials used on the top, bottom, and sides of the box.
Let's assume the width of the box is "w" meters. According to the information given, the length of the box is twice the width, so the length would be "2w" meters.
The dimensions of the box will be:
Width (w) = w meters
Length (l) = 2w meters
Height (h) = ?
The volume of the box can be calculated as length times width times height:
Volume (V) = l × w × h = (2w) × w × h = 2w^2h
Now, let's consider the costs of the materials:
Cost of top and four sides = 3 times Cost of bottom
The cost is directly proportional to the area, so we'll calculate the areas of each surface:
Area of bottom surface = length (l) × width (w) = (2w) × w = 2w^2
Area of top surface = length (l) × width (w) = (2w) × w = 2w^2
Area of four side surfaces = height (h) × width (w) × 4 = h × w × 4
Now, let's find the equation for the total cost (C) in terms of the width (w) and height (h):
C = 3 × (Cost of bottom surface) + Cost of top surface + Cost of four side surfaces
C = 3 × (2w^2) + 2w^2 + (h × w × 4)
Since we want to find the most economic dimensions, we need to minimize the cost. To do that, we differentiate the cost equation with respect to either width (w) or height (h) and equate it to zero. In this case, let's differentiate with respect to width (w).
dC/dw = 12w + 4w - 8w = 0
16w = 8w
w = 0
Since we obtain a contradiction when setting the derivative equal to zero, it implies that there is no minimum or maximum for the cost equation.
However, we can consider practical constraints for the width and height of the box. For example, the width can be assumed to be a positive value, and the height can be assumed to be greater than zero.
Therefore, we can conclude that the most economical dimensions for the box are when the width (w) is a positive value, and the height (h) is greater than zero.
w * 2w * h = c
So, h = c/(2w^2)
So if the bottom costs $1/m^2, the total cost is
C(w,h) = w*2w + 3*w*2w + 3*2(wh + 2wh)
= 7w^2 + 18wh
C(w) = 7w^2 + 18w*c/(2w^2)
= 7w^2 + 9c/w
dC/dw = 14w - 9c/w^2
dC/dw=0 when (14w^3-9c) = 0
That is, when w = ∛(9c/14)
and so on to the length and height.