A cylindrical tank is to have a capacity of 1000 m³. It is to fit into a foundry that is 12 m wide with a height of 11 m.
The base of the tank will cost half as much as the top. The metal for the side of the tank will cost four fifths as much as the top.
An equation that can be used to model the cost of the tank is C(r) = (3/2) pi(r)^2 + 1600/r. Determine any restrictions on r. You need not solve further.
obviously 2 r </= 12
volume = pi r^2 h =1000
then pi r^2 = 1000/h
but h</= 11
so pi r^2 >/= 1000/11
Well, well, well, it seems we have some math and geometry going on here. Let's unravel this puzzle together!
We have a cylindrical tank with a capacity of 1000 m³. That's one big tank! Now, we need to find an equation to model the cost of the entire tank. To do that, we need to consider the base, the sides, and the top of the tank. Let me break it down for you:
The base of the tank will cost half as much as the top. So, if we let the cost of the top be "T" (let's say it's in bucks), then the cost of the base will be T/2. Easy peasy!
Now, let's move on to the sides. The metal for the side will cost four fifths as much as the top. So, the cost of the sides will be (4/5)T. Got it?
To model the cost of the whole tank, we can add up the costs of the base, sides, and top. Let's call this cost function C and the radius of the tank r (in meters):
C(r) = T/2 + (4/5)T + (3/2)πr²
But wait, we have a constraint here! The tank needs to fit into the foundry, which is 12 m wide with a height of 11 m. This means that the radius of the tank cannot exceed half of the width (6m) and the tank's height cannot exceed the height of the foundry (11m). Therefore, we have:
0 < r ≤ 6 and 0 < C(r) ≤ 11
And there you have it! We've set up the equation to model the cost of the tank, and we've found the restrictions on the radius. Now, it's time to crunch the numbers (or let someone else do it). Good luck!
To determine the restrictions on the radius (r), we need to consider the given dimensions of the foundry.
First, let's consider the width of the foundry, which is 12 m. Since the tank needs to fit within this width, the diameter of the tank must be less than or equal to 12 m.
Since the diameter is equal to twice the radius, we have the inequality: 2r ≤ 12
Dividing both sides of the inequality by 2, we get: r ≤ 6
Therefore, the radius of the tank must be less than or equal to 6 m in order to fit within the foundry.
To determine any restrictions on r, we need to analyze the given equation C(r) = (3/2)π(r)^2 + 1600/r.
In this equation, C(r) represents the cost of the tank, and r represents the radius of the tank.
Let's break down the equation:
(3/2)π(r)^2 represents the cost of the base and top of the tank. The base will cost half as much as the top, so we can assume that half of the cost is allocated for the base and half for the top. Therefore, (3/2)π(r)^2 represents the cost for both the base and top.
1600/r represents the cost of the side of the tank. It is mentioned that the metal for the side costs four-fifths as much as the top. Thus, 1600/r represents the cost for the side.
Now, to determine restrictions on r, we need to consider the following:
1. The radius (r) cannot be negative because it represents a physical dimension (distance). Therefore, r ≥ 0.
2. The radius (r) cannot be zero because that would represent a point instead of a cylindrical tank, which doesn't make sense in this context. Therefore, r > 0.
3. There might be additional restrictions depending on the specific context of the problem, such as manufacturing constraints or practical limitations. However, these additional constraints are not stated in the given information.
Therefore, the restrictions on r are r > 0.