Answer:
The dimensions of the pencil holder that have a volume of 9π and are constructed for the cheapest cost are a radius of 3 units and a height of 1 unit.
Explanation:
Let's begin by using the given information to write equations for the volume and cost of the pencil holder.
Since the pencil holder is a right circular cylinder without a top, its volume can be calculated using the formula:
V = πr^2h
where V is the volume, r is the radius of the circular base, and h is the height of the cylinder. We are given that the volume is 9π, so we can substitute this value and simplify the equation:
9π = πr^2h
r^2h = 9
We are also told that the cost of the side is 3/8 of the cost of the bottom. Let's call the cost of the bottom "B" and the cost of the side "S". The cost of the pencil holder is the sum of the cost of the bottom and the cost of the side, so we can write:
Cost = B + S
We want to minimize the cost of the pencil holder, so we need to express the cost of the side in terms of the cost of the bottom. Since the side is 3/8 of the cost of the bottom, we can write:
S = (3/8)B
Now we can substitute this expression for S into the equation for the cost:
Cost = B + S
Cost = B + (3/8)B
Cost = (11/8)B
So the cost of the pencil holder is (11/8) times the cost of the bottom.
To minimize the cost, we want to find the dimensions of the pencil holder that satisfy the volume equation while minimizing the cost equation. We can use the volume equation to solve for one of the variables in terms of the other, and then substitute that expression into the cost equation. We can then find the minimum cost by taking the derivative of the cost equation with respect to the remaining variable and setting it equal to zero.
Let's solve the volume equation for h in terms of r:
r^2h = 9
h = 9/r^2
Now we can substitute this expression for h into the cost equation:
Cost = (11/8)B
Cost = (11/8)(r^2B + (3/8)rB)
We can simplify this equation by factoring out rB:
Cost = (11/8)rB(r + 3/8)
Now we can take the derivative of the cost equation with respect to r:
dCost/dr = (11/8)B(r + 3/8) - (11/8)rB
dCost/dr = (11/64)B(24 - 8r)
Setting this equal to zero and solving for r, we get:
(11/64)B(24 - 8r) = 0
24 - 8r = 0
r = 3
So the radius of the circular base is 3 units. Now we can use the volume equation to solve for the height:
9π = πr^2h
9π = π(3)^2h
h = 1
So the height of the cylinder is 1 unit.
Therefore, the dimensions of the pencil holder that have a volume of 9π and are constructed for the cheapest cost are a radius of 3 units and a height of 1 unit.