A square-based, box-shaped shipping crate is designed to have a volume of 16ft^3. The material used to make the base costs twice as much (per ft^2) as the material in the sides, and the material used to make the top costs half as much (per ft^2) as the material in the sides. What are the dimensions of the crate that minimize the cost of materials?

To find the dimensions that minimize the cost of materials, we need to consider the cost of each component and the volume constraint.

Let's denote the side length of the square base as x, and the height of the box as h.

Since the volume of the box is given as 16 ft^3, we have the equation:

Volume = x^2 * h = 16 ft^3 (Equation 1)

Now, we need to express the cost of materials in terms of x and h.

Let's assume the cost of the material for the sides is C. Therefore, the cost of the material for the base is 2C, and the cost of the material for the top is (1/2)C.

The total cost of the materials is then given by the equation:

Cost = (2C * x^2) + (C * x^2) + ((1/2)C * x^2) = (5/2)C * x^2 (Equation 2)

To minimize the cost, we can differentiate the cost equation with respect to x and set it equal to zero:

dCost/dx = (5/2)C * 2x = 0

Solving this equation, we find x = 0.

However, this value of x does not make physical sense for the dimensions of the crate.

Therefore, we need to consider the constraint equation (Equation 1):

x^2 * h = 16

We can solve this equation for h by substituting x = 16 / x^2:

h = 16 / x^2

Substituting this value of h into the cost equation (Equation 2), we get:

Cost = (5/2)C * x^2 = (5/2)C * (16 / x^2) = (40C) / x^2

To find the minimized cost, we need to minimize this equation.

Since the cost is inversely proportional to x^2, we can say that the minimized cost occurs when x^2 is maximized.

Considering the constraint equation (Equation 1), we find that x^2 = 16 / h.

Substituting this value into the cost equation, we get:

Cost = (40C) / (16 / h) = (40C * h) / 16 = (5C * h) / 2

Now, to minimize this equation, we need to minimize h. From Equation 1, we know that x^2 * h = 16. Since x^2 and h are both positive, we can conclude that minimizing h occurs when x^2 is minimized.

Therefore, to minimize the cost of materials, we need to minimize both x^2 and h. Based on the constraint equation (Equation 1), the only possibility is when x = h = √16 = 4 ft.

Thus, the dimensions of the crate that minimize the cost of materials are:
- The side length of the square base (x) = 4 ft
- The height of the box (h) = 4 ft

To minimize the cost of materials, we need to find the dimensions of the crate that minimize the surface area, as the cost of materials is directly proportional to the surface area. Let's denote the length of each side of the base of the crate as "x" and the height as "h".

The volume of the box is given as 16ft^3, so we have the equation:

Volume = length * width * height
16 = x * x * h

Furthermore, we can express the surface area of the box in terms of x and h:

Surface Area = 2 * (area of the base) + (area of the sides) + (area of the top)
Surface Area = 2(x * x) + 4(x * h) + (x * x)

Now, let's consider the costs of the materials. The cost of the base material (per ft^2) is twice that of the material used in the sides, and the cost of the top material (per ft^2) is half that of the sides.

Let's denote the cost per square foot of the material used in the sides as "c". Therefore, the cost per square foot of the base material is "2c", and the cost per square foot of the top material is "c/2".

Now, we can express the total cost of materials as:

Cost = (cost per ft^2 of the base material) * (area of the base) + (cost per ft^2 of the side material) * (area of the sides) + (cost per ft^2 of the top material) * (area of the top)

Cost = (2c) * (x * x) + (c) * (4(x * h)) + (c/2) * (x * x)

To minimize the cost, we need to minimize the surface area of the box.

To proceed, we can substitute the volume equation (16 = x * x * h) into the surface area equation to express the surface area in terms of a single variable (x):

Surface Area = 2(x * x) + 4(x * h) + (x * x) = 3(x * x) + 4(x * h)

Since we have h in terms of x from the volume equation, we can substitute it into the surface area equation:

Surface Area = 3(x * x) + 4(x * (16 / (x * x))) = 3(x * x) + 64 / x

Now, to find the value of x that minimizes the surface area, we can differentiate the surface area equation with respect to x and set it equal to zero:

d(Surface Area) / dx = 6x - 64 / (x^2) = 0

To solve this equation, let's multiply through by x^2 to eliminate the denominator:

6x^3 - 64 = 0

Now we can solve for x:

6x^3 = 64
x^3 = 64/6
x^3 = 32/3

Taking the cube root of both sides:

x = (32/3)^(1/3)

Now that we have the value of x, we can substitute it back into the volume equation to find h:

16 = x * x * h
16 = ((32/3)^(1/3)) * ((32/3)^(1/3)) * h
16 = (32/3)^(2/3) * h

Now solve for h:

h = 16 / (32/3)^(2/3)

Hence, the dimensions of the crate that minimize the cost of materials are:

Length (x) ≈ (32/3)^(1/3)
Height (h) ≈ 16 / (32/3)^(2/3)

Note: The values of x and h are approximations since they involve a cube root calculation.

Since the base is square, and the volume is constant, there is only one variable, such as the side of the base, x.

The height is therefore h=V/x².

The cost of the base is
Cb=C1*x²
Cost of the four sides is
Cs=C2*(4x*h)
Cost of the cover is
Cc=C3*x²

Total cost as a function of x
C(x)= C1*x²+C2*(4x*V/x²)+C3*x²
=(C1+C3)x²+ 4C2*V/x

Differentiate with respect to x and equation f'(x) to zero. Solve for x0 where the cost is maximum/minimum.

To find out if f(x0) is a maximum or minimum, calculate the second derivative and evaluate f"(x0). If f"(x0)>0, f(x0) is a minimum. If f"(x0)<0, f(x0) is a maximum.

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