A silo is to be made in the form of a cylinder surmounted by a hemisphere. The cost of construction per square foot of surface area is twice as great for the hemisphere as for the cylinder. Determine the dimensions to be use (in terms of V) if the volume is fixed at V and its cost of construction is to be minimum. Neglect the thickness of the silo and waste in construction.

I assume we are not worried about the base of the silo.

So we need the "sleeve" of the silo plus a hemisphere

let the radius be of the hemisphere be r ft and the height of the cylinder part be h ft

surface area (SA) = 2πrh + (1/2)4πr^2
Cost = 1(2πrh) + 2(1/2)4πr^2
= 2πrh + 4πr^2
but πr^2 h + (1/2)(4/3)πr^3 = V , where V is a constant
h = (V - (2/3)πr^3)/(πr^2)
sub that back into Cost = ....
simplify, then take the derivative
set that equal to zero and solve for r
Make sure you treat V as a constant.

Ah, constructing a silo, huh? Well, I guess it's time for some clown math, buckle up!

Let's start by breaking down the problem. We have a cylinder surmounted by a hemisphere. Since the volume is fixed, we can represent it as follows:

V = Volume of cylinder + Volume of hemisphere

Now, let's say the radius of the cylinder is r and the radius of the hemisphere is also r (because symmetry is good for the soul). The height of the cylinder doesn't matter in terms of the volume, so we'll represent it with a variable, let's call it h.

The volume of the cylinder can be calculated as Vcylinder = πr^2h, and the volume of the hemisphere, being half a sphere, is Vhemisphere = (2/3)πr^3.

Since we know the costs per square foot are twice as much for the hemisphere as for the cylinder, we can represent the total cost as:

Cost = Cost of cylinder surface area + Cost of hemisphere surface area

Now, let's calculate the surface areas. The surface area of a cylinder is Acylinder = 2πrh, and the surface area of a hemisphere is Ahemisphere = 2πr^2 (because the hemisphere is only half of a sphere).

Plugging all that into the equation for cost, we get:

Cost = (2πrh)(Cost of cylinder) + (2πr^2)(Cost of hemisphere)

Now, since we want to minimize the cost, let's take the derivative of the cost with respect to r, and set it equal to zero:

d(Cost)/dr = 2πh(Cost of cylinder) + 4πr(Cost of hemisphere) = 0

Simplifying that equation gives us:

h(Cost of cylinder) + 2r(Cost of hemisphere) = 0

Now, we can rearrange that equation to solve for h:

h = -(2r)(Cost of hemisphere)/(Cost of cylinder)

Again, since we want to minimize the cost, let's find the critical points of h. To do that, let's assume that V, Cost of cylinder, and Cost of hemisphere are all positive constants.

Since h must be positive, it means that (2r)(Cost of hemisphere) must be negative. This implies that Cost of hemisphere must be negative, which doesn't make any sense. So, our assumption is wrong.

Therefore, there are no critical points for h. This means that the value of h doesn't matter in terms of minimizing the cost.

So, we can conclude that the dimensions to be used (in terms of V) for minimum cost are:

- The height of the cylinder can be any positive number.
- The radius of the cylinder and the hemisphere must be equal.

And that's it! That's the clown-tastic solution to your silo construction problem. I hope you enjoyed the chuckles along the way!

To determine the dimensions that will minimize the cost of construction, we need to express the cost in terms of a single variable and then find the minimum point.

Let's start by expressing the volume of the silo in terms of one variable. The silo consists of a cylinder and a hemisphere. The volume of a cylinder is given by the formula Vcylinder = πr^2h, where r is the radius of the cylinder and h is the height. The volume of a hemisphere is given by the formula Vhemisphere = (2/3)πr^3, where r is the radius of the hemisphere.

Since both the cylinder and hemisphere have the same radius, let's call it r. The height of the cylinder portion, h, will be the variable we will express in terms of V.

The total volume of the silo is given by V = Vcylinder + Vhemisphere, so we have V = πr^2h + (2/3)πr^3.

To express the cost of construction, we need to determine the surface area of the silo. The surface area of the cylinder is given by Acylinder = 2πrh, and the surface area of the hemisphere is given by Ahemisphere = 2πr^2.

The total surface area of the silo is Atotal = Acylinder + Ahemisphere, so we have Atotal = 2πrh + 2πr^2.

The cost of construction per square foot of surface area is twice as great for the hemisphere as for the cylinder. Let's assume the cost per square foot for the cylinder is Ccylinder. Then, the cost per square foot for the hemisphere is 2Ccylinder.

The total cost of construction, Ctotal, is given by Ctotal = Ccylinder * Acylinder + 2Ccylinder * Ahemisphere.

Substituting the expressions for the surface areas, we have Ctotal = Ccylinder * (2πrh) + 2Ccylinder * (2πr^2).

To determine the dimensions that minimize the cost, we will express Ctotal in terms of one variable. Since we want to express everything in terms of the fixed volume V, we need to eliminate the variable r.

From the volume equation, V = πr^2h + (2/3)πr^3, we can solve for h: h = (V - (2/3)πr^3) / πr^2.

Substituting this expression for h back into the total cost equation, we have Ctotal = Ccylinder * (2πr((V - (2/3)πr^3) / πr^2)) + 2Ccylinder * (2πr^2).

Simplifying, we get Ctotal = 2Ccylinder * (V - (2/3)πr^3) / r + 4Ccylinder * πr^2.

To find the minimum point of Ctotal, we need to take the derivative with respect to r and set it equal to zero.

dCtotal/dr = -4Ccylinder * (2πr^2) / r^2 + 8Ccylinder * πr = 0.

Simplifying, we get -8Ccylinder + 8Ccylinder * r^3 = 0.

Dividing by 8Ccylinder and rearranging, we have r^3 = 1.

Taking the cube root of both sides, we get r = 1.

We have found that the radius of the cylinder and hemisphere should be 1. Now let's find the corresponding height, h.

From the volume equation, V = π(1^2)h + (2/3)π(1^3), we have V = πh + (2/3)π, or V - (2/3)π = πh.

Simplifying, we get h = (V - (2/3)π) / π.

So, the dimensions of the silo that will minimize the cost of construction are a radius of 1 and a height of (V - (2/3)π) / π.

To determine the dimensions that minimize the cost of construction, let's break down the problem into steps:

Step 1: Understand the problem
We need to find the dimensions of the silo that result in the minimum cost of construction, given that the volume is fixed.

Step 2: Define the variables
Let's assume the radius of the cylinder is r and the height of the cylinder is h. The radius of the hemisphere is also r, as it is surmounted on the cylinder.

Step 3: Determine equations and constraints
We know that the volume of a cylinder is given by V = πr^2h, and the volume of a hemisphere is V_hemisphere = (2/3)πr^3. Since the total volume is fixed at V, we can equate these two equations:
V = πr^2h + (2/3)πr^3

Step 4: Simplify the equation
V = πr^2h + (2/3)πr^3
V = πrh(r + (2/3)r^2)

Step 5: Determine the cost equations
The cost equation for the cylinder is C_cylinder = πr^2 + 2πrh. Since the cost per square foot of surface area is twice as great for the hemisphere, the cost equation for the hemisphere is C_hemisphere = 2πr^2 + 4πr^2.

Step 6: Simplify the cost equations
C_cylinder = πr^2 + 2πrh
C_hemisphere = 2πr^2 + 4πr^2
C_hemisphere = 6πr^2

Step 7: Determine the total cost equation
The total cost equation is defined as C_total = C_cylinder + C_hemisphere.

C_total = πr^2 + 2πrh + 6πr^2
C_total = πr^2(1 + 6) + 2πrh

Step 8: Define the objective function
Our objective is to minimize the cost, so we have to minimize C_total.

Step 9: Apply constraints
Since the volume is fixed at V, we can substitute the volume equation obtained in Step 4 into the cost equation obtained in Step 7.

C_total = πr^2(1 + 6) + 2πrh
C_total = 7πr^2 + 2πrh
C_total = 7πr^2 + 2πrh
Substituting V = πrh(r + (2/3)r^2), we get
C_total = 7πr^2 + 2πrh
C_total = 7V/(πr) + 2πrh

Step 10: Minimize the cost
To minimize the cost, we take the derivative of the cost equation with respect to r and set it to zero to find the critical points:

dC_total/dr = -7V/(πr^2) + 2πh = 0

Solving for h:
h = 7V/(2πr^2)

Substituting this into the volume equation:
V = πr^3 + (2/3)πr^3

Simplifying:
V = (5/3)πr^3

Solving for r:
r = (3V/(5π))^(1/3)

Substituting r back into the equation for h:
h = 7V/(2πr^2)

Step 11: Finalize the dimensions in terms of V
The dimensions to be used are:
Height of the cylinder, h = 7V/(2π[(3V/(5π))^(1/3)])^2
Radius of the cylinder = Radius of the hemisphere = r = (3V/(5π))^(1/3)