To find the minimum amount of metal sheet required in the construction of the vessel, let's start by determining the necessary dimensions of the cylindrical tank.
Let's assume the radius of the circular base of the tank is 'r' cm and the height of the tank is 'h' cm.
The volume of the cylindrical tank is given by the formula:
Volume = πr^2h
Given that the volume of the tank is 8000 cm^3, we can set up the equation:
8000 = πr^2h
Now, we need to minimize the surface area of the cylindrical tank. The total surface area of the tank, including the two circular bases and the curved surface, is given by the formula:
Surface Area = 2πr^2 + 2πrh
We need to minimize this surface area given the constraint that the volume is 8000 cm^3.
To minimize a function subject to a constraint, we can use the method of Lagrange multipliers.
Let L be the Lagrangian function defined as:
L(r, h, λ) = 2πr^2 + 2πrh + λ(8000 - πr^2h)
To minimize the surface area subject to the volume constraint, we need to find the critical points of L. This can be done by taking the partial derivatives of L with respect to r, h, and λ, and setting them equal to zero:
∂L/∂r = 4πr + 2πhλ = 0
∂L/∂h = 2πr + πr^2λ = 0
∂L/∂λ = 8000 - πr^2h = 0
Simplifying these equations, we get:
2r + hλ = 0 (equation 1)
r + r^2λ = 0 (equation 2)
8000 - πr^2h = 0 (equation 3)
From equation 3, we can solve for h:
h = 8000/(πr^2)
Substituting this value of h into equation 1, we get:
2r + (8000/(πr^2))λ = 0
Simplifying this equation, we have:
2r^3 + (8000/π)λ = 0 (equation 4)
Substituting the value of h from equation 3 into equation 2, we get:
r + r^2λ = 0
Simplifying this equation, we have:
r(rλ + 1) = 0
This equation gives us two critical points: r = 0 and rλ + 1 = 0. However, r = 0 is not a valid solution as we need a non-zero radius.
Therefore, we have rλ + 1 = 0, which implies rλ = -1, or λ = -1/r.
Substituting this value of λ into equation 4, we get:
2r^3 - (8000/π)(1/r) = 0
Multiplying through by r and rearranging, we have:
2r^4 - (8000/π) = 0
Simplifying further, we get:
r^4 = (4000/π)
Taking the fourth root of both sides, we get:
r = (4000/π)^(1/4)
Now that we have the value of r, we can substitute it back into equation 3 to find h:
8000 - πr^2h = 0
8000 - π[(4000/π)^(1/4)]^2h = 0
Simplifying, we have:
8000 - π(4000/π)^(1/2)h = 0
8000 - (4000/π)^(1/2)h = 0
Subtracting 8000 from both sides and rearranging, we get:
(4000/π)^(1/2)h = 8000
h = 8000/[(4000/π)^(1/2)]
Now, we have the values of r and h, we can calculate the surface area:
Surface Area = 2πr^2 + 2πrh
Substituting the values of r and h, we get:
Surface Area = 2π[(4000/π)^(1/4)]^2 + 2π[(4000/π)^(1/4)](8000/[(4000/π)^(1/2)])
Simplifying, we have:
Surface Area = 2π[(4000/π)^(1/2)] + 2π(8000/[(4000/π)^(1/4)])
Surface Area = 2π[(4000/π)^(1/2)] + 2π(8000/[(4000/π)^(1/4)])
To find the minimum amount of metal sheet required in the construction of the vessel, we calculate the surface area using the above equation.
Please note that the numerical value of the surface area will be involved, and it depends on the precise value of π (pi). Therefore, we need to calculate the value of π to the desired level of accuracy in order to obtain the specific numerical answer.
A cylindrical tank closed at both ends with circular base is to hold 8000cm^3 of liquid. Find the minimum amount of metal sheet required in the construction of the vessel
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