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Chapter 6: Problem 21

A closed rectangular container with a square base is to have a volume of \(2000 \mathrm{cm}^{3} .\) It costs twice as much per square centimeter for the top and bottom as it does for the sides. Find the dimensions of the container of least cost.

Short Answer

Expert verified
The container's dimensions are 10 cm by 10 cm by 20 cm.

Step by step solution

01

Define Variables

Let the side length of the square base be \( x \) cm and the height of the container be \( h \) cm. The volume constraint is \( x^2 \cdot h = 2000 \).
02

Express Cost Function

The surface area for the top and bottom is \( 2x^2 \), and for the sides is \( 4xh \). If the cost for the sides is \( c \) per cm虏, the cost for the top and bottom is \( 2c \) per cm虏. Total cost is thus \( 2c \cdot 2x^2 + c \cdot 4xh = 4cx^2 + 4cxh \).
03

Substitute Height from Volume Constraint

From \( x^2h = 2000 \), express \( h \) in terms of \( x \): \( h = \frac{2000}{x^2} \). Substitute \( h \) into the cost function to get \( 4cx^2 + \frac{8000c}{x} \).
04

Differentiate Cost Function

Differentiate the cost function \( 4cx^2 + \frac{8000c}{x} \) with respect to \( x \): \( \frac{d}{dx}(4cx^2) + \frac{d}{dx}\left(\frac{8000c}{x}\right) = 8cx - \frac{8000c}{x^2} \).
05

Find Critical Points

Set the derivative equal to zero to find critical points: \( 8cx - \frac{8000c}{x^2} = 0 \). Solve for \( x \) to get \( 8x^3 = 8000 \). Thus, \( x^3 = 1000 \), so \( x = 10 \).
06

Determine the Height

Substitute \( x = 10 \) back into the volume constraint \( x^2h = 2000 \): \( 100h = 2000 \), so \( h = 20 \).
07

Confirm Minimum Cost

Check the second derivative (\( 8c + \frac{16000c}{x^3} \)) to confirm it is positive at \( x = 10 \), ensuring the cost is minimized.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Constraints
When dealing with optimization problems involving volume, we often encounter volume constraints that limit the size or dimensions of the object we're optimizing. In this scenario, we have a closed rectangular container with a square base, and the volume is fixed at 2000 cm鲁.
To handle this:
  • Define variables to represent the dimensions, such as let the side length of the square base be \(x\) and the height of the container be \(h\).
  • The volume constraint can be expressed by the equation \(x^2 \, h = 2000\). This relationship ensures that the volume remains constant while you adjust the dimensions to minimize cost.
Notably, by rearranging this equation, one can express one dimension in terms of the other, which aids in further calculations like minimizing cost. For instance, solving for \(h\) gives \(h = \frac{2000}{x^2}\). This handy form is pivotal in further steps like setting up the cost function.
Cost Function
Designing a cost function is crucial in optimization problems since it guides finding the optimal solution under given constraints. For our container, the cost depends on the area of the materials used.
Understanding the cost function involves:
  • Identifying that the top and bottom surfaces of the container cost twice as much per cm虏 as the sides.
  • The surface area of the top and bottom is \(2x^2\), while the surface area of the sides is \(4xh\).
  • The total cost then incorporates these areas: \(2c \, 2x^2 + c \, 4xh = 4cx^2 + 4cxh\), where \(c\) is the cost per cm虏 for the sides.
  • By substituting the expression for \(h\) derived from the volume constraint (\(h = \frac{2000}{x^2}\)), the cost function becomes: \(4cx^2 + \frac{8000c}{x}\).
The cost function thus links the dimensions and the costs, serving as the target for optimization to achieve minimal cost.
Differentiation
Differentiation is a powerful mathematical tool employed to find points of interest such as maxima, minima, or points of inflection by analyzing a function's derivative. In optimizing the cost of the container:
  • The cost function \(4cx^2 + \frac{8000c}{x}\) is differentiated with respect to \(x\). This step gives the derivative \(8cx - \frac{8000c}{x^2}\).
  • Setting this derivative equal to zero reveals the critical points, essential for potential minima or maxima: \(8cx - \frac{8000c}{x^2} = 0\).
  • Solving this equation yields \(x^3 = 1000\), and thus \(x = 10\).
  • To confirm that this \(x\) value indeed gives a minimum cost, the second derivative test is employed: \(8c + \frac{16000c}{x^3}\). If positive, as checked at \(x = 10\), it indicates a local minimum.
Hence, differentiation aids in pinpointing the optimal container dimensions that satisfy the constraints while minimizing costs.

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Most popular questions from this chapter

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