To solve this problem, we need to use calculus and the method of optimization. Here are the steps:

1. First, let's denote the side length of the square base as x (in cm) and the height of the box as h (in cm). The volume V of the box is given by V = x^2 * h. We know that V must be 32,000 cm^3, so we have the equation x^2 * h = 32,000.

2. We want to minimize the amount of material used, which is the same as minimizing the surface area A of the box. The surface area is given by A = x^2 + 4*x*h (the area of the base plus the area of the four sides).

3. We can express h in terms of x using the volume equation: h = 32,000 / x^2. Substituting this into the surface area equation gives A = x^2 + 4*x*(32,000 / x^2) = x^2 + 128,000 / x.

4. To find the minimum of A, we take the derivative of A with respect to x and set it equal to zero. The derivative of A is A' = 2x - 128,000 / x^2. Setting this equal to zero gives 2x = 128,000 / x^2, or x^3 = 64,000. Taking the cube root of both sides gives x = 40 cm.

5. Substituting x = 40 cm into the equation for h gives h = 32,000 / 40^2 = 20 cm.

So the dimensions that minimize the amount of material used are a base side length of 40 cm and a height of 20 cm.

Question

To solve this problem, we need to use calculus and the method of optimization. Here are the steps:

1. First, let's denote the side length of the square base as x (in cm) and the height of the box as h (in cm). The volume V of the box is given by V = x^2 * h. We know that V must be 32,000 cm^3, so we have the equation x^2 * h = 32,000.

2. We want to minimize the amount of material used, which is the same as minimizing the surface area A of the box. The surface area is given by A = x^2 + 4*x*h (the area of the base plus the area of the four sides).

3. We can express h in terms of x using the volume equation: h = 32,000 / x^2. Substituting this into the surface area equation gives A = x^2 + 4*x*(32,000 / x^2) = x^2 + 128,000 / x.

4. To find the minimum of A, we take the derivative of A with respect to x and set it equal to zero. The derivative of A is A' = 2x - 128,000 / x^2. Setting this equal to zero gives 2x = 128,000 / x^2, or x^3 = 64,000. Taking the cube root of both sides gives x = 40 cm.

5. Substituting x = 40 cm into the equation for h gives h = 32,000 / 40^2 = 20 cm.

So the dimensions that minimize the amount of material used are a base side length of 40 cm and a height of 20 cm.

Knowee AI · Accepted Answer

To solve this problem, we need to use calculus and the method of optimization. Here are the steps:

1. First, let's denote the side length of the square base as x (in cm) and the height of the box as h (in cm). The volume V of the box is given by V = x^2 * h. We know that V must be 32,000 cm^3, so we have the equation x^2 * h = 32,000.

2. We want to minimize the amount of material used, which is the same as minimizing the surface area A of the box. The surface area is given by A = x^2 + 4*x*h (the area of the base plus the area of the four sides).

3. We can express h in terms of x using the volume equation: h = 32,000 / x^2. Substituting this into the surface area equation gives A = x^2 + 4*x*(32,000 / x^2) = x^2 + 128,000 / x.

4. To find the minimum of A, we take the derivative of A with respect to x and set it equal to zero. The derivative of A is A' = 2x - 128,000 / x^2. Setting this equal to zero gives 2x = 128,000 / x^2, or x^3 = 64,000. Taking the cube root of both sides gives x = 40 cm.

5. Substituting x = 40 cm into the equation for h gives h = 32,000 / 40^2 = 20 cm.

So the dimensions that minimize the amount of material used are a base side length of 40 cm and a height of 20 cm.

A box with a square base and open top must have a volume of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.sides of base cmheight cm

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