A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $8 per square meter for the base and $4 per square meter for the sides, what is the cost of the least expensive tank?
If we let the length be L and the height be H, then we have
4LH = 36
LH = 9
L = 9/H
The cost
C = 4L*8 + 4*4H*4
C = 32(9/H) + 64H
C = 288/H + 64H
dC/dH = 64 - 288/H^2
dC/dH = 0 when H = √2.5 = 2.12
so L = 4.24
C = 288/2.12 + 64*2.12 = $271.53
To find the cost of the least expensive tank, we need to determine the dimensions of the tank that minimize the cost.
Let's denote the length of the tank as L, and the height as H. Since the width is given as 4 meters, we have the dimensions of the tank as: width = 4m, length = L, and height = H.
The volume of a rectangular tank is given by the formula V = length × width × height. In this case, we have V = 36 cubic meters.
Substituting the given values, we get:
36 = L × 4 × H
9 = L × H
To minimize the cost, we need to minimize the surface area of the tank. The surface area is the sum of the areas of the base and the four sides.
The area of the base is given by the formula A_base = length × width = L × 4.
The area of each side is given by the formula A_side = width × height = 4 × H.
The total surface area, A_total, is given by:
A_total = A_base + 2 × A_side
= L × 4 + 2 × (4 × H)
= 4L + 8H
To minimize the cost, we need to minimize the surface area by finding the values of L and H that satisfy the volume constraint.
Substituting the equation 9 = L × H into the surface area equation, we get:
A_total = 4L + 8H
= 4(9/H) + 8H
= 36/H + 8H
To minimize the cost, we need to find the values of H that minimize A_total. To find this minimum, we can take the derivative of A_total with respect to H and set it equal to zero:
d(A_total)/dH = -36/H^2 + 8 = 0
Solving this equation, we find:
36/H^2 = 8
H^2 = 36/8
H^2 = 9/2
H = sqrt(9/2)
H = 3/sqrt(2)
Since the height cannot be negative, we take H = 3/sqrt(2).
Substituting this value back into the equation 9 = L × H, we can solve for L:
9 = L × (3/sqrt(2))
L = 9 * sqrt(2) / 3
L = 3 * sqrt(2)
Therefore, the dimensions of the tank that minimize the cost are: width = 4m, length = 3 * sqrt(2)m, and height = 3/sqrt(2)m.
Now, we can calculate the cost of the least expensive tank by multiplying the surface area by the respective costs:
Cost = [4L + 8H] × $8 (for the base) + [2 × (4H)] × $4 (for the sides)
Cost = [4 * (3 * sqrt(2)) + 8 * (3/sqrt(2))] * 8 + [2 * (4 * (3/sqrt(2)))] * 4
Cost = [12 * sqrt(2) + 24/sqrt(2)] * 8 + [24/sqrt(2)] * 4
Calculating this expression will give you the final answer, which is the cost of the least expensive tank.