Problem 30
Question
A closed box in the form of a rectangular parallelepiped with a square base is to have a given volume. If the material used in the bottom costs \(20 \%\) more per square inch than the material in the sides, and the material in the top costs \(50 \%\) more per square inch than that of the sides, find the most economical proportions for the box.
Step-by-Step Solution
Verified Answer
The box's economical proportions are \( x = \sqrt[3]{\frac{4V}{2.7}} \) and \( h = \frac{V}{x^2} \).
1Step 1: Define Variables
Let the side of the square base be \( x \) inches, the height of the box be \( h \) inches, and the volume of the box be \( V \) cubic inches. We have \( V = x^2h \).
2Step 2: Express Cost Functions
Let \( C_s \) be the cost per square inch of the side material. The cost per square inch for the bottom is \( 1.2C_s \) and for the top is \( 1.5C_s \).
3Step 3: Calculate Surface Area and Total Cost
Surface area of the box = bottom area \( x^2 \) + top area \( x^2 \) + side area \( 4xh \). Total cost function: \( C = 1.2C_sx^2 + 1.5C_sx^2 + 4C_sxh \), simplifying gives \( C = C_s (2.7x^2 + 4xh) \).
4Step 4: Substitute for Height using Volume
From \( V = x^2h \), solve for \( h \): \( h = \frac{V}{x^2} \). Substitute \( h \) into the cost equation: \( C = C_s (2.7x^2 + \frac{4V}{x}) \).
5Step 5: Differentiate the Cost Function
Take the derivative of \( C \) with respect to \( x \): \( \frac{dC}{dx} = 2.7C_sx - \frac{4VC_s}{x^2} \).
6Step 6: Find Critical Points by Setting the Derivative to Zero
Set \( \frac{dC}{dx} = 0 \): \( 2.7x = \frac{4V}{x^2} \). Solve for \( x \): \( x^3 = \frac{4V}{2.7} \). Simplify: \( x = \sqrt[3]{\frac{4V}{2.7}} \).
7Step 7: Calculate the Proportions
Substitute \( x \) back in \( h = \frac{V}{x^2} \) to get \( h \). Using \( x = \sqrt[3]{\frac{4V}{2.7}} \), \( h = \frac{V}{\left(\sqrt[3]{\frac{4V}{2.7}}\right)^2}\).
8Step 8: Interpretation of the Result
Therefore, the most economical proportions for the box are: square base side \( x = \sqrt[3]{\frac{4V}{2.7}} \) and height \( h = \frac{V}{x^2} \).
Key Concepts
Derivative ApplicationsCost FunctionsCritical PointsVolume and Surface Area
Derivative Applications
Understanding derivative applications is crucial in solving optimization problems, such as finding the most economical dimensions for a box. In calculus, derivatives help determine how a function changes at any given point. By finding the derivative of the cost function with respect to the variable of interest (in this case, the side of the box), we can identify where the function increases or decreases.
Using derivatives, especially in financial contexts, can significantly reduce costs by helping find the optimal point, which in calculus is often a minimum or maximum. Here, we want to minimize the cost, so we take the derivative of the cost function and set it to zero to find critical points.
These critical points will tell us where our function has potential minima or maxima, aiding in determining the most cost-effective design for our box.
Using derivatives, especially in financial contexts, can significantly reduce costs by helping find the optimal point, which in calculus is often a minimum or maximum. Here, we want to minimize the cost, so we take the derivative of the cost function and set it to zero to find critical points.
These critical points will tell us where our function has potential minima or maxima, aiding in determining the most cost-effective design for our box.
Cost Functions
Cost functions allow us to model the total expense incurred in making decisions, such as choosing the dimensions of materials for a box. By determining the cost per unit area for different materials of the box and relating them to their respective areas, we can generate a cost function for the entire box.
Here, each part of the box has a different cost per unit area: the bottom is 20% more expensive, and the top is 50% pricier than the sides. This is captured in the cost function:
Here, each part of the box has a different cost per unit area: the bottom is 20% more expensive, and the top is 50% pricier than the sides. This is captured in the cost function:
- Cost for the bottom: \(1.2C_sx^2\)
- Cost for the top: \(1.5C_sx^2\)
- Cost for the sides: \(4C_sxh\)
Critical Points
Critical points are where the derivative of a function is zero or undefined. These points are crucial in identifying where a function may have its maximum or minimum values, which is key in optimization problems.
To find these points, we took the derivative of the cost function \( \frac{dC}{dx} = 2.7C_sx - \frac{4VC_s}{x^2} \) and set it equal to zero. Solving \( 2.7x = \frac{4V}{x^2} \) allows us to calculate potential minimum points in the cost function by solving for \(x\).
After finding \(x\) as \(x = \sqrt[3]{\frac{4V}{2.7}}\), we can then verify whether these points minimize the cost by assessing the nature of the critical point through second derivatives or other methods to ensure they indeed provide a cost-effective solution.
To find these points, we took the derivative of the cost function \( \frac{dC}{dx} = 2.7C_sx - \frac{4VC_s}{x^2} \) and set it equal to zero. Solving \( 2.7x = \frac{4V}{x^2} \) allows us to calculate potential minimum points in the cost function by solving for \(x\).
After finding \(x\) as \(x = \sqrt[3]{\frac{4V}{2.7}}\), we can then verify whether these points minimize the cost by assessing the nature of the critical point through second derivatives or other methods to ensure they indeed provide a cost-effective solution.
Volume and Surface Area
Combining concepts of volume and surface area is key to solving this exercise. The volume constraint was given, \(V = x^2h\), which we utilized to link height and the base of the box. This relationship enabled us to express one variable in terms of the other.
The surface area formula, which includes the areas of the top, bottom, and sides, allowed us to create the cost function based on these areas and their respective costs. By ensuring that any changes in dimensions respected the constant volume, we could determine the dimensions that minimized the cost.
Thus, understanding how to express the height in terms of the base dimension \(h = \frac{V}{x^2}\) played a pivotal role, along with evaluating the surface area when calculating the total cost through the derived cost function. This methodology allowed for the establishing of optimal values of \(x\) and \(h\) within the problem's constraints.
The surface area formula, which includes the areas of the top, bottom, and sides, allowed us to create the cost function based on these areas and their respective costs. By ensuring that any changes in dimensions respected the constant volume, we could determine the dimensions that minimized the cost.
Thus, understanding how to express the height in terms of the base dimension \(h = \frac{V}{x^2}\) played a pivotal role, along with evaluating the surface area when calculating the total cost through the derived cost function. This methodology allowed for the establishing of optimal values of \(x\) and \(h\) within the problem's constraints.
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