Optimization: cost of materials | Applications of derivatives | AP Calculus AB | Khan Academy
TLDRThe video script explores the problem of minimizing the cost of a rectangular open-top storage container with a fixed volume of 10 cubic meters. The base's length is twice its width, and the cost of materials for the base and sides are $10 and $6 per square meter, respectively. The script guides through setting up the cost function with respect to the width (x) and height (h), expressing h as a function of x using the volume constraint. It then finds the critical points by taking the derivative of the cost function and applying the second derivative test to confirm a minimum value. The minimum cost is found to be approximately $163.54 for a container with dimensions based on the critical point value of x, which is approximately 1.65 meters for the width, making the length 3.3 meters, and the height just under 2 meters.
Takeaways
- 😀 The problem involves finding the cost-optimal dimensions for a rectangular storage container with an open top, given specific cost constraints for materials.
- 😮 The volume of the container must be 10 cubic meters.
- 😕 The length of the base of the container is twice its width.
- 😊 The cost of material for the base is $10 per square meter, while material for the sides costs $6 per square meter.
- 😅 The cost function for the container is derived as a function of its dimensions, with the aim of minimizing cost.
- 🤔 The cost function depends on both the width (x) and height (h) of the container, making optimization a multi-variable problem.
- 🧐 To optimize, the height (h) is expressed as a function of the width (x) using the volume constraint.
- 🔍 Critical points of the cost function are found by taking its derivative and setting it equal to zero.
- 😌 One legitimate critical point is identified at approximately x = 1.65, likely corresponding to the minimum cost.
- 🎉 The minimum cost of the container is calculated to be approximately $163.54, with dimensions derived from the optimal width.
Q & A
What is the volume requirement for the open-top rectangular storage container?
-The volume of the storage container must be 10 cubic meters.
What is the relationship between the length and width of the base of the container?
-The length of the base is twice the width. If we denote the width as 'x', then the length is '2x'.
What are the costs per square meter for the base and sides of the container?
-The material for the base costs $10 per square meter, and the material for the sides costs $6 per square meter.
What is the formula for calculating the cost of the container as a function of x (width) and h (height)?
-The cost of the container is calculated as 20x^2 (cost of the base) + 12xh (cost of two side panels) + 24xh (cost of the other two side panels), which simplifies to 20x^2 + 36xh.
How is the height 'h' of the container expressed as a function of 'x' (width) using the volume constraint?
-The height 'h' can be expressed as a function of 'x' by rearranging the volume formula (2x^2 * h = 10) to solve for 'h', which gives h = 5 / x^2.
What is the simplified cost function of the container after substituting h with its expression in terms of x?
-After substituting h = 5 / x^2 into the cost function, the simplified cost function becomes 20x^2 + 180x^(-1).
How do you determine the critical points of the cost function?
-To find the critical points, you take the derivative of the cost function with respect to 'x', set it equal to zero, and solve for 'x'. The points where the derivative is zero or undefined are the candidate critical points.
What is the critical point for the width 'x' that minimizes the cost of the container?
-The critical point for the width 'x' that minimizes the cost is approximately x = (9/2)^(1/3), which is approximately 1.65 meters.
How do you confirm that the critical point found is a minimum value rather than a maximum?
-You use the second derivative test. If the second derivative of the cost function at the critical point is positive, it indicates that the function is concave upwards at that point, which means it is a minimum value.
What is the approximate cost of the material for the cheapest container when the width 'x' is 1.65 meters?
-The approximate cost of the material for the cheapest container is $163.54.
What are the approximate dimensions of the cheapest container in terms of width, length, and height?
-With the width 'x' being approximately 1.65 meters, the length would be twice that, around 3.3 meters. The height 'h' is approximately 5 / (1.65^2), which is a little under two meters.
What does the final cost calculation take into account?
-The final cost calculation takes into account the material costs for the base and all four sides of the container, as well as the dimensions that yield the minimum cost while meeting the volume constraint.
Outlines
📏 Designing the Cheapest Open-Top Storage Container
This paragraph introduces the problem of finding the cheapest way to construct an open-top storage container with a volume of 10 cubic meters. The base's length is twice its width, and the material costs are specified for the base ($10 per square meter) and the sides ($6 per square meter). The goal is to minimize the cost of materials while adhering to these constraints. The paragraph also includes a visual representation of the container and the mathematical setup to calculate the cost as a function of the width (x) and height (h).
🧮 Calculating the Cost Function and Relating Width to Height
The second paragraph focuses on establishing a mathematical relationship between the width (x), length (2x), and height (h) of the container to satisfy the volume requirement of 10 cubic meters. By doing so, it finds an expression for the height in terms of x, which is h = 5/x^2. This allows the paragraph to express the total cost of the container as a function of x alone, incorporating both the cost of the base and the sides. The paragraph concludes by differentiating this cost function to find critical points, which are necessary for determining the minimum cost.
📐 Applying the Second Derivative Test to Find the Minimum Cost
In the final paragraph, the second derivative test is used to confirm that the critical point found in the previous paragraph indeed corresponds to a minimum value for the cost function. This is done by evaluating the second derivative at the critical point and showing that it is positive, indicating concavity upwards and thus a minimum. With the confirmation that the critical point corresponds to the minimum cost, the paragraph then calculates the cost of the material for the cheapest container by substituting the critical value of x into the cost function, resulting in an approximate cost of $163.54 for the container.
Mindmap
Keywords
💡Volume
💡Base
💡Length and Width
💡Cost
💡Optimization
💡Derivative
💡Second Derivative Test
💡Critical Points
💡Height (h)
💡Side Panels
💡Concave Upwards
Highlights
The goal is to find the cost of materials for the cheapest open-top rectangular storage container with a volume of 10 cubic meters.
The base length is twice the width (length = 2x, width = x).
Material for the base costs $10 per square meter, while the sides cost $6 per square meter.
The cost of the container is a function of the width (x) and height (h), with the base and sides having different material costs.
The area of the base is calculated as width times length (x * 2x), resulting in a cost of $10 times x squared.
The cost of the sides involves two different dimensions: x times h for the shorter sides and 2x times h for the longer sides.
The total cost function is derived as 20x squared plus 36xh, representing the cost as a function of x and h.
To optimize the cost function with respect to x, the height (h) must be expressed as a function of x using the volume constraint.
The volume constraint leads to h = 5 / x squared, which is then substituted into the cost function.
The optimized cost function becomes 20x squared plus 180x to the power of -1, simplifying the problem to a single variable optimization.
The critical points of the cost function are found by setting its first derivative to zero and solving for x.
The critical point x = (9/2)^(1/3) ≈ 1.65 is identified, indicating a potential minimum cost.
The second derivative test confirms that x = 1.65 is indeed a minimum point, ensuring the cost function is concave upwards at this value.
Substituting x = 1.65 into the cost function yields an approximate minimum cost of $163.54 for the container.
The dimensions of the cheapest container are approximately 1.65 meters in width, 3.3 meters in length, and just under 2 meters in height.
The material cost for the cheapest container is determined to be around $163.54, highlighting the efficiency of the optimization process.
The problem-solving approach demonstrates the application of calculus in optimizing real-world design problems, such as minimizing material costs for containers.
Transcripts
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