Your iron works has been contracted to design and build a 500-ft3 square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding half-inch-thick stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weight as little as possible. What dimensions do you tell the shop to use?
clearly the weight of the box is a function of the material used, which in turn depends on the surface area of the sheet metal used.
So we want a minimum surface area.
let each side of the base be x ft, and let the height be y ft.
Area = x^2 + 4xy
but x^2 y = 500
y = 500/x^2
area = x^2 + 4x(500/x^2) = x^2 + 2000/x
d(area)/dx = 2x -2000/x^2 = 0 for a min of area
2x = 2000/x^2
x^3 = 1000
x = 10 , then y = 500/10^2 = 5
the box should have a base of 10 ft by 10 ft, and a height of 5 ft.
Well, as the Clown Bot, I must say, let's think outside the box for this tank design! And by outside the box, I mean outside the heavy weight!
To make the tank weight as little as possible, we should focus on minimizing the amount of steel used. So, let's make it as skinny as possible but still functional, like a supermodel tank!
Since this is a square-based tank, let's go for square dimensions. We want to minimize the area of the steel plates, so we need to find the smallest possible square base that can hold 500 cubic feet of paper company goodness.
To calculate the base area, we need to find the square root of 500, which equals approximately 22.36 feet. Since we want a square base, both sides would be 22.36 feet.
Now, let's consider the height. We want to minimize the amount of steel used on the sides, so we'll go for a relatively short height. Let's say 3 feet.
So, the dimensions I suggest for this tank are a base of 22.36 ft by 22.36 ft and a height of 3 ft. This way, we reduce the weight by using minimum materials while still keeping it functional. Happy tank tinkering!
To find the dimensions that will make the tank weight as little as possible, we can use the principles of optimization.
Let's start by determining the objective function, which is the weight of the tank. The weight can be calculated by multiplying the density of stainless steel with the volume of the tank.
Next, we need to establish the constraints for the dimensions of the tank. We know that the tank is square-based, meaning that the base is a square. The volume of a square-based tank is given by multiplying the area of the base (side^2) with the height.
Now, let's denote the side length of the square base as "s" and the height as "h". We can write the volume equation as:
Volume = s^2 * h = 500 ft^3
To minimize the weight, we can differentiate the weight equation with respect to one of the variables (s or h) and find where the derivative is equal to zero.
Before we proceed with the optimization, we need to consider the welding requirements. Since half-inch-thick stainless steel plates are being used and they need to be welded along their edges, the dimensions of the tank should be such that plates can be easily welded together.
Considering these considerations, a practical approach would be to keep the dimensions of the tank as close to a cube as possible. Therefore, we choose the side length "s" to be equal to the height "h".
Now, let's solve the volume equation:
s^2 * s = 500 ft^3
s^3 = 500 ft^3
s = ∛(500)
s ≈ 7.37 ft (approximately)
Thus, the dimensions to tell the shop to use for the tank are approximately:
Base(side length) = 7.37 ft
Height = 7.37 ft
These dimensions will make the tank weigh as little as possible while considering structural integrity and the welding requirements.
To design the holding tank with the least weight possible, we need to consider the concept of surface area and volume. A larger surface area will result in more welding and a heavier tank, while a larger volume will also increase the weight of the tank due to the extra material needed.
Let's break down the problem step by step:
1. Determine the known variables:
We are given the volume of the tank, which is 500 ft^3. This means the tank can hold 500 cubic feet of liquid or material.
2. Determine the unknown variables:
We need to find the dimensions of the base and height that will minimize the weight of the tank. Let's assume the base of the tank is a square, with sides labeled 'x', and the height of the tank is labeled 'h.'
3. Formulate the equations:
Since the base is square-shaped, the surface area of each side will be x^2. Thus, the total surface area of the base would be 4x^2.
The surface area of the rectangular sides can be determined by multiplying the height 'h' by the perimeter of the base square ('4x'). Hence, the total surface area of the sides would be 4xh.
To minimize the weight, we want to minimize the total surface area.
4. Create the objective function:
We can now create an objective function that represents the total surface area of the tank, and we need to minimize this function:
Total Surface Area (A) = Base Area + Side Area = 4x^2 + 4xh
5. Apply constraints:
We also need to consider that the volume of the tank must be 500 ft^3, thus:
Base Area * Height = Volume
x^2 * h = 500
6. Solve the problem:
To minimize the weight of the tank, we will differentiate the objective function A with respect to 'h,' treating 'x' as a constant:
dA/dh = 4x
To find the values of 'x' and 'h' that minimize the weight, we need to set the derivative equal to zero:
4x = 0
Since 'x' cannot be zero (as it represents the base side length), this implies that the derivative is always positive.
Hence, there is no minimum value or critical point, and we conclude that the tank's weight will be minimized when the sides have an infinitesimally small value, i.e., a height as close to zero as possible.
Therefore, the dimensions that will make the tank weigh as little as possible are a square base with a side length as large as possible (while still meeting the volume requirement), and a height as small as possible.