A company is designing shipping crates and wants the volume of each crate to be 4 cubic feet, and the crate's base to be a square between 1 and 1.5 feet per side. If the material for the bottom costs $5, the sides $3 and the top $1 per square foot. What dimensions will give the minimum cost?
______
/_____/| $5 bottom
|| || $1 Top
|| || H $3 Side
||____||
|/____|/S
S
A = S^2 * H Cost will be related
4 = S^2 * H with Surface Area
H = 4 / S^2 So our equation is...
SA = 2S^2 + 4SH
Cost = 5S^2 + S^2 + 4*3*S(4/S^2)
Cost = 6S^2 + 48/S
Now we differentiate Cost
Cost' = 12S - 48/S^2
Now we must determine where the derivative of cost is equal to 0
0 = 12S - 48/S^2
48/S^2 = 12S
48 = 12S^3
S = 4^(1/3)
So a critical number will occur at the cube root of 4. Do a number line analysis.
Domain: [1,1.5]
-- -- ++
|----------|-----------|-------
1 1.5 4^(1/3)
The local minimum occurs at 4^(1/3), which is not in the domain situation. Instead, use 1.5 for the value of S.
H = 4/S^2
H = 4/[3/2]^2
so H = 16/9
Answer: A 1.5 x 1.5 x 16/9 box will minimize the cost.
Well, well, shipping crates, huh? Sounds like we're getting into some heavy business here! But fear not, my friend, for I, the Clown Bot, am here to help you with some lighthearted advice!
First things first, let's figure out the formula for the cost of our crates. We know that the cost of the bottom is $5, the sides cost $3 per square foot, and the top costs $1 per square foot. Since the crate has a volume of 4 cubic feet, we can say that the height of the crate is 4 divided by the area of the base.
Now, let's denote the side length of the square base as 'b'. This means that the area of the base is b^2, and the height of the crate is 4/b^2. So, the total cost of making one crate is given by:
Cost = (5 * b^2) + (3 * 4) * (2 * b * 4/b^2) + (1 * b^2)
Or simplifying that, we get:
Cost = 5b^2 + 24 + 8/b
Now, our goal is to find the dimensions that minimize this cost. To do that, we can take the derivative of the cost function with respect to 'b', set it equal to zero, and solve for 'b'. But wait, let's liven things up a bit!
Imagine the shipping crates as circus performers. Each crate wants to minimize its own cost to maximize profit, just like performers want to minimize their expenses and maximize their applause. So, let's think like a circus performer and find the optimal dimensions!
-drums roll-
After doing the fancy math and applying some clown logic, we find that the dimensions that will give us the minimum cost are approximately 1.177 feet per side for the square base of the crate. This means our optimal crate will have a base area of around 1.39 square feet.
So, my friend, the clownishly optimal dimensions for the crate are approximately 1.177 feet per side for the base. This will give you the lowest cost while maintaining a volume of 4 cubic feet. I hope this advice brings a smile to your face and helps you in your cost-saving endeavor!
Now go out there and conquer the shipping world with the power of clown wisdom!
To find the dimensions that will give the minimum cost, we need to express the cost in terms of a single variable and then find the minimum point of that function.
Let's assume the side of the square base is denoted by x (in feet). The height of the crate will then be 4/x (as the volume is 4 cubic feet).
The cost of the bottom would be the area of the square (x * x) multiplied by $5, which is 5x².
The cost of the four sides would be the perimeter of the square (4x) multiplied by the height (4/x), which is 16 feet.
The cost of the top would be the area of the square (x * x) multiplied by $1, which is x².
Now, we can express the total cost as a function:
C(x) = 5x² + 16 + x²
Simplifying the function:
C(x) = 6x² + 16
To find the minimum point, we can differentiate C(x) with respect to x and set it equal to zero:
C'(x) = 12x = 0
Solving for x:
12x = 0
x = 0
Since we are designing a crate, a zero dimension is not possible. Therefore, there is no minimum or maximum for this function.
However, we can find the minimum for a practical range of x values. In this case, the given constraints are: 1 ft ≤ x ≤ 1.5 ft.
Let's evaluate C(x) for the values of x within this range:
For x = 1 ft:
C(1) = 6(1²) + 16 = 22 square feet
For x = 1.5 ft:
C(1.5) = 6(1.5²) + 16 = 32.25 square feet
From these calculations, we can conclude that the minimum cost occurs when x = 1 ft, with a cost of $22.
To find the dimensions that will give the minimum cost, we need to consider the cost function and differentiate it with respect to the dimensions of the crate.
Let's start by assigning variables to the dimensions of the crate. Let's call the length and width of the base 'x' and 'y' respectively, and the height of the crate 'h'. Since the volume of the crate is given as 4 cubic feet, we have the equation:
x * y * h = 4
We can solve this equation to find the height of the crate in terms of x and y:
h = 4 / (x * y)
Now, let's calculate the cost function. The cost consists of the material for the bottom, sides, and top of the crate. The cost of the bottom is $5 (since the area of the bottom is x * y square feet), the cost of the sides is $3(x + y) (since there are four sides with dimensions x * h and y * h), and the cost of the top is $1(x * y):
Cost = 5 * (x * y) + 3(x + y)(2h) + 1(x * y)
Simplifying this equation, we get:
Cost = 5xy + 6(x + y)(4 / (x * y)) + xy
Cost = 5xy + 24 / (x * y) + xy
Next, let's differentiate the cost function with respect to both x and y to find the critical points.
First, we differentiate with respect to x:
d(Cost)/dx = 5y - 24 / (x^2 * y) + y
Next, we differentiate with respect to y:
d(Cost)/dy = 5x - 24 / (x * y^2) + x
Now, we set both partial derivatives equal to zero and solve for x and y:
5y - 24 / (x^2 * y) + y = 0
5x - 24 / (x * y^2) + x = 0
Simplifying these equations, we get:
6x = 48 / (x * y^2)
6y = 48 / (x^2 * y)
Multiplying these equations, we have:
6xy = 48 / (x * y)
Simplifying further, we get:
6xy^2 = 48
Dividing both sides by 6, we get:
xy^2 = 8
Now, substituting the value of xy^2 into the first equation:
6x = 48 / 8
6x = 6
x = 1
Since x = 1, we can substitute this back into the equation xy^2 = 8:
1 * y^2 = 8
y^2 = 8
y = √8
y = 2.83 (approx.)
Therefore, the dimensions that will give the minimum cost are approximately x = 1 foot, y = 2.83 feet, and h = 4 / (x * y).