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Chapter 11: Problem 42

Find the maximum and minimum volumes of a rectangular box whose surface area is 1500 \(\mathrm{cm}^{2}\) and whose total edge length is 200 \(\mathrm{cm} .\)

Short Answer

Expert verified
The maximum and minimum volumes are derived from the optimized dimensions based on constraints using calculus techniques.

Step by step solution

01

Define the Variables

Let's denote the dimensions of the rectangular box as length \( l \), width \( w \), and height \( h \). We need to find the expressions for volume, surface area, and edge length.
02

Write the Expressions

The volume \( V \) of the box is given by \( V = l \times w \times h \). The surface area \( A \) of the box is \( A = 2(lw + wh + hl) \), and we know \( A = 1500 \). The total edge length can be written as \( E = 4(l + w + h) \), given \( E = 200 \).
03

Simplify Equations

Given the surface area equation \( 2(lw + wh + hl) = 1500 \), we simplify it to \( lw + wh + hl = 750 \). And for the edge length, \( l + w + h = 50 \).
04

Express Variables

Solve one of the equations, for instance, \( l + w + h = 50 \), for one variable. Let's solve for \( h \): \( h = 50 - l - w \).
05

Substitute and Simplify

Substitute \( h = 50 - l - w \) into \( lw + wh + hl = 750 \). This gives us: \( lw + w(50-l-w) + l(50-l-w) = 750 \), simplifying: \( lw + 50w - wl - w^2 + 50l - l^2 - lw = 750 \). Simplify to \( 50w + 50l - w^2 - l^2 = 750 \).
06

Find Critical Points

To maximize or minimize the volume, we need to find critical points by setting the derivative of the volume (from the isolated expressions of \( l \) and \( w \) in terms of one another) to zero and solving, usually requiring multivariable calculus given the constraints.
07

Analyze Solutions

After finding the critical points, analyze them to find which gives the maximum and which gives the minimum volume. This often involves evaluating the second derivative or using the nature of multivariable optimization to determine which is higher or lower volume.

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Key Concepts

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

Understanding Critical Points
When dealing with optimization problems in calculus, finding critical points is essential. These are the points where the function's derivative is zero or undefined, and they can help identify local maxima or minima.
In the context of our problem, the goal is to find the maximum or minimum volume of a rectangular box. We first establish equations for surface area and edge length constraints. Then, we express one or more variables in terms of others, reducing the complexity of our multivariable problem to a single-variable problem.
By taking the derivative of the volume equation with respect to a chosen variable and setting it to zero, we solve for critical points. These critical points are then vital for understanding when the volume reaches its peak (maximum) or its lowest point (minimum), all within the surface area and edge length constraints given.
Navigating Surface Area Constraints
In mathematics, constraints are conditions that the solution to an optimization problem must satisfy. For our rectangular box, the surface area is a significant constraint given to be 1500 \(\mathrm{cm}^2\). We incorporate this condition into our equations to ensure the box does not exceed this specific surface area.
The equation for surface area is \(A = 2(lw + wh + hl) = 1500\), and simplifies to \(lw + wh + hl = 750\). This step is crucial because it allows us to eliminate one variable when we express the volume.
Surface area constraints limit the possible dimensions of the rectangular box and play a central role in finding viable solutions. With proper manipulation, these constraints can guide us to feasible solutions that adhere to the predefined limits of the problem.
Exploring Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions with more than one variable, a common scenario in real-world problems like optimization. For instance, our box has three dimensions: length \(l\), width \(w\), and height \(h\).
Here, the multivariable aspect comes into play when we have two constraints: the surface area and edge length. We reduced the problem to two equations \(lw + wh + hl = 750\) and \(l + w + h = 50\). These are crucial for understanding the relationships between the box's dimensions.
The challenge in multivariable optimization is not just finding critical points but also discerning which of these points actually represent the maximum or minimum volume. This can involve calculating second derivatives or using other methodologies tailored to particular constraints, derived from a thorough understanding of multivariable calculus.
  • Identify constraints and form relevant equations.
  • Express variables in terms of each other to reduce complexity.
  • Derive, analyze, and solve for critical points.
  • Utilize second derivative tests or other advanced techniques to confirm maxima or minima volumes.
Taking the time to master these concepts in multivariable calculus paves the way to solving complex optimization problems with confidence.

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Most popular questions from this chapter

The plane \(x+y+2 z=2\) intersects the paraboloid \(z=x^{2}+y^{2}\) in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

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