A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must have a volume of 1600 cubic feet. The hemispherical ends cost twice as much per square foot of surface area as the sides. Find the dimensions that will minimize the cost. (Round your answers to three decimal places.)

To minimize the cost, we need to find the dimensions of the tank that will minimize the surface area since the cost is directly proportional to the surface area.

Let's consider the dimensions of the tank:
- Let the radius of the cylinder be 'r'.
- The height of the cylinder is 'h'.
- The radius of the hemisphere is also 'r'.

To find the surface area, we need to consider two parts:
1. The surface area of the cylindrical part (excluding the bases): This can be calculated as the lateral surface area of a cylinder, which is given by A_lateral = 2πrh.
2. The surface area of the two hemispheres (including the bases): This can be calculated as the surface area of two hemispheres, which is given by A_hemisphere = 2πr^2.

The total surface area of the tank is the sum of the two parts:
A_total = A_lateral + A_hemisphere

Now, we have two constraints:
1. The volume of the tank is 1600 cubic feet: This constraint can be expressed as V = A_base * h = 1600, where A_base is the area of the base of the cylinder, given by A_base = πr^2.
2. The cost per square foot of the hemisphere is double that of the sides: This can be expressed as Cost_hemisphere = 2 * Cost_sides.

To minimize the cost, we need to minimize the cost function:
Cost = Cost_sides * (A_lateral) + Cost_hemisphere * (A_hemisphere)

Now, we have two cost components: Cost_sides and Cost_hemisphere. Let's denote them as C_s and C_h, respectively.

So, the cost function becomes:
Cost = C_s * (2πrh) + C_h * (2πr^2)

Given that C_h = 2 * C_s, we can rewrite the cost function as:
Cost = C_s * (2πrh + 4πr^2)

Now, we need to express the cost function solely in terms of one variable to find its minimum value using calculus. Let's express it in terms of h only:
Cost = C_s * (2πrh + 4πr^2)
= C_s * 2πr (h + 2r)

Since V = A_base * h = πr^2 * h = 1600, we can express h in terms of r:
h = 1600 / (πr^2)

Substituting this value of h back into the expression of the cost function:
Cost = C_s * 2πr (h + 2r)
= C_s * 2πr (1600 / (πr^2) + 2r)
= C_s * (3200 / r + 4πr^2)

Now, we have the cost function expressed solely in terms of one variable, r. To minimize the cost, we need to find the derivative of the cost function with respect to r, set it equal to zero, and solve for r. This will give us the critical points, and we can then determine if they are minimums.

Differentiating the cost function with respect to r gives us:
d(Cost)/dr = C_s * (3200 / r^2 + 8πr)

Setting this derivative equal to zero and solving for r:
0 = C_s * (3200 / r^2 + 8πr)

Dividing both sides by C_s to simplify:
0 = 3200 / r^2 + 8πr

Multiplying through by r^2 to eliminate the fraction:
0 = 3200 + 8πr^3

Rearranging the equation:
8πr^3 = -3200

Dividing both sides by 8π:
r^3 = -400 / π

Taking the cube root of both sides:
r = (-400 / π)^(1/3)

Now that we have the value of r, we can substitute it back into the equation for h to find its value:
h = 1600 / (πr^2)
h = 1600 / (π * [(-400 / π)^(1/3)]^2)

Now, you can calculate the values of r and h, and then substitute them into the cost function to find the minimum cost.