Problem 18 - Contemporary Precalculus

Question

An open-top box with a square base is to be constructed from 120 square centimeters of material. What dimensions will produce a box (a) of volume 100 cubic centimeters? (b) with largest possible volume?

Step-by-Step Solution

Answer

Answer: (a) For a box with a volume of 100 cubic centimeters, the dimensions are approximately 4.58 cm by 4.58 cm by 4.75 cm. (b) For the largest possible volume, the dimensions are approximately sqrt(40) cm by sqrt(40) cm by 30/sqrt(40) cm.

Step 1: Write the equation for the volume

The volume of a box is given by the product of its length, width, and height. Since the base of the box is square, the dimensions are x, x, and y. Therefore, the volume (V) is: V = x^2*y

Step 2: Write the constraint equation

The total material used to construct the box is given as 120 square centimeters. The area of the square base is x^2, and the area of the four sides is 4*y*x. Therefore, we have the following constraint equation: x^2 + 4*y*x = 120

Step 3: Solve the constraint equation for y

Solve the constraint equation for y in terms of x: y = (120 - x^2) / (4*x)

Step 4: Substitute y in terms of x into the volume equation

Replace y in the volume equation with the expression found in step 3: V(x) = x^2 * (120 - x^2) / (4*x)

Step 5: Simplify the volume equation

Simplify the volume equation by cancelling x: V(x) = (120x - x^3)/4 (a) To find the dimensions that will produce a box of volume 100 cubic centimeters:

Step 6: Set the volume equation equal to 100

Set V(x) = 100 and solve for x: (120x - x^3)/4 = 100

Step 7: Solve for x

Solve the equation for x: x^3 - 120x + 400 = 0 The solution to the above equation can be found numerically, and we get x ≈ 4.58 cm

Step 8: Find the value of y

Substitute the value of x found in step 7 into the expression for y in terms of x: y ≈ (120 - 4.58^2) / (4*4.58) ≈ 4.75 cm For a box with a volume of 100 cubic centimeters, the dimensions are approximately 4.58 cm by 4.58 cm by 4.75 cm. (b) To find the dimensions that will produce the largest possible volume:

Step 9: Find the critical points of the volume equation

Differentiate V(x) with respect to x: V'(x) = (120 - 3x^2)/4 Set V'(x) = 0 and solve for x: 120 - 3x^2 = 0 x^2 = 40 x = ±sqrt(40) Since x must be positive, we ignore the negative solution and consider x = sqrt(40).

Step 10: Check for the global maximum

Analyze the second derivative of V(x) to check if the critical point we found is a global maximum: V''(x) = -6x/4 Substitute the value of x found in step 9 and determine the sign of V''(sqrt(40)): V''(sqrt(40)) = -6(sqrt(40))/4 < 0 Since the second derivative is negative at the critical point, it is a global maximum.

Step 11: Find the dimensions

Substitute the value of x found in step 9 into the expression for y in terms of x: y = (120 - (sqrt(40))^2) / (4*sqrt(40)) = 30/sqrt(40) The dimensions that will produce the largest possible volume are approximately sqrt(40) cm by sqrt(40) cm by 30/sqrt(40) cm.

Key Concepts

Volume Calculations

When dealing with optimization problems, particularly in geometry, understanding volume calculations is crucial. The volume of a three-dimensional object is simply the measure of the amount of space it occupies. In our problem, we are dealing with an open-top box with a square base. The volume of this box depends on its base area and height.
For a square base with side length \( x \) and height \( y \), the volume \( V \) is calculated as:

• \( V = x^2\cdot y \)

This reflects how the base area \( x^2 \) is multiplied by the height \( y \) to determine the overall space enclosed by the box. By mastering this basic calculation, you set the foundation for solving more complex aspects of the problem.

Constraint Equations

In optimization problems, constraint equations provide the conditions that solutions must satisfy. Here, we need to construct the box using a specific amount of material, 120 square centimeters. This means that both the base and the sides of the box must add up to this given material area.
The constraint equation relates the dimensions of the box (\( x \) and \( y \)) to the available material:

• \( x^2 + 4yx = 120 \)

This equation ensures that the sum of the area of the square base \( x^2 \) and the areas of the four vertical sides \( 4yx \) remains 120 square centimeters. It is this balance between the dimensions and the constraint that guides our optimization efforts moving forward.

Derivative Applications

In calculus, derivatives play a key role in understanding how functions change and in solving optimization problems. Derivatives help us determine the rate of change of a function, which is essential in finding maximum or minimum values.
To optimize the volume of the box, we first derive the expression for volume \( V(x) \) with respect to \( x \):

• \( V(x) = \frac{120x - x^3}{4} \)

By taking the derivative \( V'(x) \), we can find changes in volume relative to changes in \( x \):

• \( V'(x) = \frac{120 - 3x^2}{4} \)

This derivative, set to zero, will help us pinpoint critical points, leading to the optimal solutions.

Critical Points in Calculus

Critical points are values in the domain of a function where its derivative is zero or undefined. These points are instrumental in finding the maxima or minima of functions, which are central to optimization.
In our example, after deriving the volume function \( V(x) \), we solved \( V'(x) = 0 \) to find the critical points:

• \( 120 - 3x^2 = 0 \)
• Solve to find \( x = \sqrt{40} \)

Since we are dealing with a tangible object, we only consider the positive solution. Once we find \( x \), a second derivative test confirms if this point indeed corresponds to a maximum or minimum. The negative second derivative indicates that \( x = \sqrt{40} \) is a point of maximum volume for the material constraint, ensuring our box is as large as can be built within the given limits of material.