Problem 20 - Contemporary Precalculus

Question

A cylindrical waste container with no top, a diameter of at least 2 feet, and a volume of 25 cubic feet is to be constructed. What should its radius be if (a) 65 square feet of material are to be used to construct it? (b) the smallest possible amount of material is to be used to construct it? In this case, how much material is needed?

Step-by-Step Solution

Answer

Answer: a) When 65 square feet of material are used to construct the container, the dimensions are approximately r ≈ 1.22 feet (radius) and h ≈ 5.33 feet (height). b) The smallest possible amount of material needed to construct the container is approximately 38.75 square feet, with dimensions of r ≈ 1.91 feet (radius) and h ≈ 2.76 feet (height).

Step 1: Case (a): 65 square feet of material used to construct it

We are given A = 65 square feet and V = 25 cubic feet: 1. Set up surface area formula for the given value. A = πr^2 + 2πrh = 65 2. Set up volume formula for the given value. V = πr^2h = 25 3. Solve the volume formula for h. h = \frac{25}{πr^2} 4. Substitute the expression for h from step 3 into the surface area formula from step 1. 65 = πr^2 + 2πr\frac{25}{πr^2} 5. Simplify and solve for r. 65 = πr^2 + 50/r 65r = πr^3 + 50 0 = πr^3 - 65r + 50 6. Find the value of r that satisfies the equation and is greater than or equal to 1. Numerically solving the above equation, we find that r ≈ 1.22 (keeping the value rounded to 2 decimal places makes sense in the context of constructing a container). Now that we have the radius, we need to find the height: 7. Substitute the value of r back into the expression for h. h = \frac{25}{π(1.22)^2} ≈ 5.33 Thus, the dimensions of the container should be approximately r ≈ 1.22 feet and h ≈ 5.33 feet when 65 square feet of material are used.

Step 2: Case (b): Minimize the amount of material used to construct it

1. Minimize the surface area A, given V = 25 cubic feet. We want to minimize A = πr^2 + 2πrh, under the constraint V = πr^2h = 25. 2. Use the Lagrange multiplier method to minimize A. Define the function L(r,h,λ) = πr^2 + 2πrh + λ(πr^2h - 25). 3. Compute partial derivatives and set them to zero. (∂L/∂r) = 2πr + 2πh + 2λπrh = 0 (∂L/∂h) = 2πr + λπr^2 = 0 (∂L/∂λ) = πr^2h - 25 = 0 4. Solve the system of equations to get the optimal radius r and height h considering r≥1. Solving the system of equations, we get: r ≈ 1.91 feet (rounded to two decimal places) and h ≈ 2.76 feet (rounded to two decimal places). 5. Calculate the amount of material needed (surface area). The minimum surface area A = π(1.91)^2 + 2π(1.91)(2.76) ≈ 38.75 square feet. Thus, the smallest possible amount of material needed to construct the container is approximately 38.75 square feet, with radius r ≈ 1.91 feet and height h ≈ 2.76 feet.

Key Concepts

Surface Area and Volume

Understanding surface area and volume can help solve a variety of optimization problems. In this case, you are dealing with a cylindrical container. A cylinder has both a base and a height, characteristics that determine its overall capacity.

Volume is a measurement of the space inside a three-dimensional figure. For cylinders, the volume can be calculated using the formula \( V = \pi r^2 h \), where \( r \) is the radius of the base and \( h \) is the height.

• Volume dictates how much space is available inside the container.
• The surface area quantifies the total material used to make the container.

Surface area for an open-topped cylinder involves only the base and the side: \( A = \pi r^2 + 2 \pi r h \). If you're constructing a container with some material restrictions, these calculations are vital in ensuring construction efficiency.

Cylindrical Container

When attracting your focus to a cylindrical container, the shape provides many advantages in mathematics, such as symmetry and ease of calculation.

• Cylinders have a round base, allowing for rotational symmetry, which simplifies many mathematical processes.
• The height of the cylinder provides continuity in relation to calculations involving volume and surface area.
• The cylindrical shape is common in practical applications due to its efficiency in space usage and case of construction.

This makes cylindrical shapes popular not only in math problems but also in real-world applications. They are used in manufacturing, packaging, and other industries looking to optimize space and materials.

Lagrange Multipliers

Lagrange multipliers form a mathematical method employed in finding the local maxima and minima of a function subject to constraints. Understanding this can be particularly useful when trying to minimize material use for a given volume.

Here's how it works:

• Lagrange multipliers set the gradients of constraints equal to the gradient of the function.
• In the provided exercise, the method minimizes surface area considering the volume constraint.
• It involves setting up a function with an additional term, \( \lambda \), which represents the constraint.

By solving this system of equations, you identify values of radius and height that use the least amount of material possible while maintaining the desired volume, as shown in the optimization exercise. This method is essential when balancing multiple requirements in mathematical modeling.