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An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the dome costs twice as much per square foot as the cylindrical wall, what are the most economical proportions for the given volume?
To find the most economical proportions for the given volume, we need to determine the dimensions of the cylinder and the hemisphere that minimize the cost.
Let's start by assigning variables to the dimensions:
- Let "r" be the radius of the cylinder base (in feet).
- Let "h" be the height of the cylinder (in feet).
Now, let's consider the volume of the observatory. The volume of the cylinder is given by V_cylinder = πr^2h, and the volume of the hemisphere is V_hemisphere = (2/3)πr^3.
The total volume V_total is the sum of the two volumes:
V_total = V_cylinder + V_hemisphere
= πr^2h + (2/3)πr^3
= (πr^2h + (2/3)πr^3).
Next, we need to express the cost in terms of the dimensions, considering that the cost of the dome is twice that of the cylindrical wall per square foot.
Let "c" be the cost per square foot for the cylindrical wall. Therefore, the cost per square foot for the dome is 2c.
Now, let's calculate the surface area of the observatory. The surface area of the cylindrical wall is given by A_cylinder = 2πrh, and the surface area of the hemisphere (dome) is A_hemisphere = 2πr^2.
The total surface area A_total is the sum of the two surface areas:
A_total = A_cylinder + A_hemisphere
= 2πrh + 2πr^2
= 2πr(h + r).
The total cost C_total is the product of the surface area and the cost per square foot:
C_total = A_total * c + A_total * 2c
= (2πr(h + r))(c + 2c)
= 2πr(h + r)(c + 2c).
Our goal is to minimize the total cost while keeping the volume V_total constant. To do this, we need to take the derivative of C_total with respect to one of the variables (r or h). However, since we have two variables, it's more convenient to express one variable in terms of the other using the volume equation.
Rearrange the volume equation as follows:
πr^2h + (2/3)πr^3 = V
πr^2h = V - (2/3)πr^3
h = (V - (2/3)πr^3) / (πr^2).
Now we can substitute the expression for "h" back into the cost equation:
C_total = 2πr(h + r)(c + 2c)
= 2πr(((V - (2/3)πr^3) / (πr^2)) + r)(c + 2c)
= 2πr((V - (2/3)πr^3 + rπr^2)) / (πr^2)(c + 2c)
= 2πr((V - (2/3)πr^3 + r^3)) / (πr^2)(c + 2c).
This equation represents the total cost in terms of r, V, and c. To find the most economical proportions, we need to take the derivative of C_total with respect to r, set it to zero, and solve for r.
Differentiating C_total with respect to r:
dC_total/dr = 2π[(V - (2/3)πr^3 + r^3)(1) - r(2/3)π(3r^2)] / (πr^2)(c + 2c)
= 2π[(V - (2/3)πr^3 + r^3) - 2(1/3)πr^3] / (πr^2)(c + 2c)
= 2π(V - (2/3)πr^3 + r^3 - (2/3)πr^3) / (πr^2)(3c)
= 2π(V + (1/3)πr^3) / (πr^2)(3c)
= 2(V + (1/3)πr^3) / (r^2)(3c).
Setting dC_total/dr to zero:
2(V + (1/3)πr^3) / (r^2)(3c) = 0.
Now, we solve for r:
V + (1/3)πr^3 = 0
V = -(1/3)πr^3
V / (-(1/3)π) = r^3
r = ((3V) / (-π))^(1/3).
Substituting this value of r back into the expression "h = (V - (2/3)πr^3) / (πr^2)", we can solve for h.
h = (V - (2/3)π((3V) / (-π))^(1/3)^3) / (π((3V) / (-π))^(1/3)^2)
= (V + 2V) / (3((3V) / (-π))^(2/3))
= (3V) / (3((3V) / (-π))^(2/3))
= (-πV) / ((3V)^(2/3)).
Thus, we have the values of r and h in terms of V, which can be considered the most economical proportions for the given volume.