A water tank is to be constructed by rotating a semi-circular trough around its diameter. Find the radius that minimizes the cost of the tank for a given volume.
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To find the radius that minimizes the cost of the tank, we need to determine the cost function and then find its minimum.
Let's start by analyzing the cost of constructing the tank. The cost is likely to be directly related to the surface area of the tank, as this is the part that would require construction materials. The volume of the tank does not come into play when determining the cost, as it only affects the dimensions and not the surface area.
The surface area of the tank consists of three parts: the inner curved surface, the outer curved surface, and the bottom.
The inner curved surface is a semicircle, and its circumference can be given by C = πd, where d is the diameter of the semicircle. Since the trough is rotated around its diameter, this circumference represents the inner curved surface.
The outer curved surface is also a semicircle with a larger radius. To determine its circumference, we can use the formula C = π(D), where D is the diameter of the outer semicircle. The cost of the outer curved surface would be higher than that of the inner curved surface because it requires more material.
The bottom of the tank is a circle with radius r, which does not change during rotation.
Since the question asks for the radius that minimizes the cost for a given volume, we need to find the cost as a function of the radius. We can write the cost function as follows:
Cost = c1 * C + c2 * C + c3 * A
Where c1, c2, and c3 are cost coefficients, C represents the costs of the inner and outer curved surfaces, and A represents the cost of the bottom circle.
The inner curved surface is a semicircle, so its circumference C1 can be written as C1 = πd = π * 2r.
Similarly, the outer curved surface is also a semicircle but with a larger diameter, so its circumference C2 can be written as C2 = π(D) = π * 2R, where R is the radius of the outer curved surface.
The area of the bottom circle can be given by A = π(r^2).
The cost function can now be rewritten as:
Cost = c1 * C1 + c2 * C2 + c3 * A
= c1 * (π * 2r) + c2 * (π * 2R) + c3 * π(r^2)
= 2π(c1r + c2R) + c3π(r^2)
Now that we have the cost function in terms of the radius, r, and the radius of the outer curve, R, we can proceed with finding the minimum cost by taking the derivative of the cost function with respect to r and setting it equal to zero.
d(Cost)/dr = 2π(c1 + c3r) = 0
Solving for r:
c1 + c3r = 0
r = -c1 / c3
Since r represents the radius, it cannot be negative. Therefore, we can discard the negative solution and conclude that the radius that minimizes the cost of the tank for a given volume is r = -c1 / c3.
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