An open-top shipping crate with square bottom and rectangular sides is to hold 32 in^3 and requires a minimum amount of cardboard. Find the most economical directions.
volume: s^2 *h where s is the bottom side length, h is height.
area: s^2+ 4sh
but 32=s^2 h so h=32/s^2
darea/ds=d(s^2 + 4*32/s)
2s-128/s^2=0
solve for s, then h
check to see that area is minimum.
To find the most economical dimensions for the open-top shipping crate, we need to minimize the amount of cardboard used.
Let's assume the side lengths of the square bottom of the crate are L inches.
The dimensions of the rectangular sides will be L * W, where W is the width of the sides in inches, and the height of the sides will be H inches.
The volume of the crate is given as 32 in^3, so we have the equation:
Volume = L * L * H = 32.
We need to find the dimensions that minimize the amount of cardboard used. The cardboard required for the bottom is L * L, and for the sides, it is 2 * (L * W) + 2 * (L * H).
To minimize the amount of cardboard used, we can use optimization techniques, such as finding the partial derivatives and setting them to zero. However, since this is a step-by-step response, we will find the dimensions through a logical approach.
1. Let's list down the factors of 32:
1, 2, 4, 8, 16, 32.
2. Since we want to minimize the amount of cardboard used, let's start by choosing the smallest factor as one of the side lengths (L). Let's start with L = 1.
Substituting L = 1 in the volume equation: 1 * 1 * H = 32.
So, H = 32.
Calculating the cardboard used:
Bottom: 1 * 1 = 1 square inch
Sides: 2 * (1 * W) + 2 * (1 * 32) = 2 * W + 64 square inches.
The total cardboard used: 1 + 2 * W + 64 = 2 * W + 65 square inches.
3. Let's calculate the total cardboard used for the other factors of 32 in the same way:
For L = 2, we get H = 16.
The total cardboard used: 4 * W + 80 square inches.
For L = 4, we get H = 4.
The total cardboard used: 8 * W + 32 square inches.
For L = 8, we get H = 1.
The total cardboard used: 16 * W + 16 square inches.
For L = 16, we get H = 0.5.
The total cardboard used: 32 * W + 8 square inches.
For L = 32, we get H = 0.25.
The total cardboard used: 64 * W + 4 square inches.
4. Comparing the total cardboard used for each factor:
L = 1: 2 * W + 65 square inches.
L = 2: 4 * W + 80 square inches.
L = 4: 8 * W + 32 square inches.
L = 8: 16 * W + 16 square inches.
L = 16: 32 * W + 8 square inches.
L = 32: 64 * W + 4 square inches.
From the above values, we can observe that the total cardboard used decreases as L increases.
5. Since we are looking for the most economical dimensions, we need to minimize the total cardboard used. As we observed, increasing L decreases the total cardboard used. Therefore, the most economical dimensions will be when L = 32.
Substituting L = 32 in the volume equation: 32 * 32 * H = 32.
So, H = 1/32.
Calculating the cardboard used:
Bottom: 32 * 32 = 1024 square inches.
Sides: 2 * (32 * W) + 2 * (32 * (1/32)) = 64 * W + 2 square inches.
The total cardboard used: 1024 + 64 * W + 2 = 64 * W + 1026 square inches.
Therefore, the most economical dimensions for the open-top shipping crate with a square bottom and rectangular sides, while holding 32 in^3, is to have a square bottom with side length of 32 inches and have the height of the sides as 1/32 inches.
To determine the most economical dimensions for the crate, we need to minimize the surface area of the cardboard used.
Let's start by assigning variables to the dimensions of the crate. Let's call the length of the sides of the square bottom "x" and the length of the rectangular sides "y" and "z".
Since the volume of the crate is given as 32 in^3, we have the following equation: x^2 * y = 32.
To find the minimum surface area, we can express it as a function of one variable. The surface area of the crate can be calculated by adding the areas of the bottom and the four sides.
The bottom area is simply x^2.
The area of each side can be calculated by multiplying the height (z) with the length (y), so the total area of all four sides is 4yz.
Therefore, the total surface area (S) is: S = x^2 + 4yz.
Now, we need to express the surface area as a function of a single variable, so we can find its minimum. To do this, we'll substitute the value of y from the volume equation.
Rearranging the volume equation, we have y = (32 / x^2).
Now, substitute this value of y in the surface area equation:
S = x^2 + 4 * (32 / x^2) * z.
To minimize the surface area, we need to find the value of x and z that minimizes the function S. To do this, we can take the derivative of S with respect to x and set it equal to zero.
dS/dx = 2x - (4 * 32 * z) / x^3.
Setting this expression equal to zero and solving for x, we have:
2x - (4 * 32 * z) / x^3 = 0.
This expression simplifies to:
2x^4 - 128z = 0.
From this equation, we find that x^4 = 64z.
Since both x and z must be positive, we can substitute x^4 = 64z back into the volume equation:
x^2 * y = 32.
(x^2) * ((32 / x^2)) = 32,
32 = 32,
which indicates that there are infinitely many solutions to this equation. Therefore, the most economical dimensions for the crate are not unique. You can choose any dimensions that satisfy the equation x^2 * y = 32, as long as they follow the condition that x^4 = 64z.
To summarize, the most economical dimensions for the open-top shipping crate with a square bottom and rectangular sides can be obtained by choosing any positive values of x, y, and z that satisfy the equation x^2 * y = 32, while also ensuring that x^4 = 64z.