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 6080 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 minimize the total surface area. Let's denote the radius of the cylinder as r and the height of the cylinder as h.
First, let's calculate the volume of the cylinder. The formula for the volume of a cylinder is given by V_cylinder = πr^2h.
Since the volume of the tank is given as 6080 cubic feet, we have:
πr^2h = 6080
Next, let's find the surface area of the cylinder. The formula for the lateral surface area of a cylinder is given by A_lateral = 2πrh.
Now, let's calculate the surface area of the cylindrical part of the tank:
A_cylinder = A_lateral = 2πrh
The total surface area of the tank is the sum of the surface area of the cylindrical part and the two hemispherical ends.
A_total = A_cylinder + 2A_hemisphere
To minimize the cost, we need to calculate the cost as a function of the surface area.
Let's assume that the cost of the sides is c dollars per square foot of surface area, then the cost of the hemispheres is 2c dollars per square foot of surface area. Therefore, the cost function is given by:
C(A_total) = c(A_cylinder) + 2c(A_hemisphere)
Since the cost function depends on the surface area, we can express it in terms of r and h.
Now, let's express A_total, A_cylinder, and A_hemisphere in terms of r and h.
For the cylindrical part:
A_cylinder = 2πrh
For each hemisphere, the surface area is given by the formula:
A_hemisphere = 2πr^2
Now, let's substitute the surface area formulas into the cost function:
C(r, h) = c(2πrh) + 2c(2πr^2)
Simplifying the cost function, we have:
C(r, h) = 2cπrh + 4cπr^2
To minimize the cost, we need to find the critical points of the cost function. This can be done by taking partial derivatives of C with respect to r and h and setting them equal to zero.
∂C/∂r = 0 and ∂C/∂h = 0
Taking the partial derivative of C with respect to r:
∂C/∂r = 2cπh + 8cπr = 0
Simplifying this equation, we have:
h + 4r = 0
Taking the partial derivative of C with respect to h:
∂C/∂h = 2cπr = 0
Simplifying this equation, we have:
r = 0
However, we are interested in finding positive values for r and h, so we'll disregard r = 0.
Substituting h = -4r into the volume equation πr^2h = 6080, we get:
πr^2(-4r) = 6080
-4πr^3 = 6080
Simplifying further, we have:
r^3 = -1520/π
Now, solving for r, we have:
r = ∛(-1520/π)
Substituting this value of r into h + 4r = 0, we get:
h - 4∛(-1520/π) = 0
Simplifying further, we have:
h = 4∛(-1520/π)
Therefore, the dimensions that will minimize the cost are:
r = ∛(-1520/π) and h = 4∛(-1520/π)
Now, substitute the value of r into the cost function to find the minimum cost:
C(r, h) = 2cπrh + 4cπr^2
Finally, substitute the values of r and h into the cost function and evaluate the result.
To find the dimensions that will minimize the cost, we need to express the cost as a function of the dimensions of the solid and then apply calculus techniques to find the minimum of that function.
Let's denote the radius of the cylindrical part of the solid as r and the height of the cylindrical part as h.
Since the solid is formed by adjoining two hemispheres to the ends of the cylinder, the total volume of the solid can be expressed as the sum of the volume of the cylinder and the volume of the two hemispheres:
Volume of the cylinder: πr^2h
Volume of one hemisphere: (2/3)πr^3 (since the volume of a hemisphere is half the volume of a sphere)
Total volume: πr^2h + 2(2/3)πr^3 = πr^2h + (4/3)πr^3
Given that the total volume must be 6080 cubic feet, we have the equation:
πr^2h + (4/3)πr^3 = 6080
To minimize the cost, we need to express the cost as a function of the dimensions. The cost consists of two components: the cost of the sides and the cost of the hemispherical ends.
Let's denote the cost per square foot of the sides as S, and since the hemispherical ends cost twice as much per square foot of surface area, the cost per square foot of the hemispheres is 2S.
The total cost (C) is given by the sum of the cost of the sides and the cost of the hemispheres:
Cost of the sides: S*(surface area of the sides)
Cost of the hemispheres: 2S*(surface area of the hemispheres)
To find the surface area of the sides, we need to find the lateral area of the cylinder, which is given by:
Surface area of the sides: 2πrh
To find the surface area of the hemispheres, we need to find the sum of the surface area of two hemispheres, which is given by:
Surface area of the hemispheres: 2*(1/2) * 4πr^2 = 4πr^2
Therefore, the total cost can be expressed as:
C = S*(2πrh) + 2S*(4πr^2)
Simplifying further:
C = 4πr^2S + 2πrSh
Now we have an expression for the cost (C) in terms of the dimensions r and h. To minimize the cost, we need to find the values of r and h that minimize this function.
To do that, we need to use the equation πr^2h + (4/3)πr^3 = 6080 as a constraint while minimizing the cost. We can rearrange this equation as:
h = (6080 - (4/3)πr^3) / (πr^2)
Now we have the height (h) in terms of the radius (r). Substituting this into the cost function, we can express the cost solely in terms of r:
C = 4πr^2S + 2πrS * [(6080 - (4/3)πr^3) / (πr^2)]
Simplifying further:
C = 4πr^2S + 2S * [(6080 - (4/3)πr^3) / r]
Now we have the cost (C) as a function of the radius (r) and the cost per square foot of the sides (S).
To find the dimensions that will minimize the cost, we need to take the derivative of the cost function with respect to r, set it equal to zero, and solve for r. Then, using the derived value of r, we can find the corresponding value of h.
Finally, we can substitute the values of r and h back into the cost function to find the minimum cost and determine the dimensions that will minimize the cost.
what, no effort since you last [posted this problem? Most of it is just geometry.
If the radius is r and the cylinder length is h, then
4/3 π r^3 + πr^2 h = 1600
so
h = (4800 - 4πr^3)/(3πr^2)
= 1600/πr^2 - 4r/3
If the sides cost $1/ft^2, then the ends cost $2/ft^2, so the cost is
c = 2*4πr^2 + 2πrh
= 2*4πr^2 + 2πr(1600/πr^2 - 4r/3)
= 3200/r + 16π/3 r^2
so, to minimize cost, find where dc/dr = 0
dc/dr = -3200/r^2 + 32πr/3
= (32πr^3-9600)/3r^2
dc/dr=0 when
32πr^3 = 9600
r^3 = 300/π
r = 4.57
h = 18.28