Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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1. Creating the Objective Function: The cost to build any section of the shed is the unit cost (2) to build a section multiplied by the number of square feet of a section. We have the unit costs, so now we need to find the square footage of each section in terms of x and/or y. a. The base: Depth = c. The roof: Depth = Length = b. The sides: 1. Left side and right side. Since these sides have the same dimensions, find the area and cost of the left side and then double it. Depth = Total Cost = Height = (cost of base) Area = 2. Front side and back side. Since these sides have the same dimensions, find the area and cost of the front side and then double it. Length = Height = Cost = Length = Area of both = Cost of both = Area of both = Cost of both Area = d. Now we can write our cost equation in terms of x and y. The objective function is the total cost to build the shed. Total cost is the cost to build the base and the sides and the roof. Since the cost to build the left and right sides are the same (and the front and back) combine those cost terms to get: Cost = (cost of left and right) (cost of front and back) (cost of roof)
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Transcribed Image Text:1. Creating the Objective Function: The cost to build any section of the shed is the unit cost (2) to build a section multiplied by the number of square feet of a section. We have the unit costs, so now we need to find the square footage of each section in terms of x and/or y. a. The base: Depth = c. The roof: Depth = Length = b. The sides: 1. Left side and right side. Since these sides have the same dimensions, find the area and cost of the left side and then double it. Depth = Total Cost = Height = (cost of base) Area = 2. Front side and back side. Since these sides have the same dimensions, find the area and cost of the front side and then double it. Length = Height = Cost = Length = Area of both = Cost of both = Area of both = Cost of both Area = d. Now we can write our cost equation in terms of x and y. The objective function is the total cost to build the shed. Total cost is the cost to build the base and the sides and the roof. Since the cost to build the left and right sides are the same (and the front and back) combine those cost terms to get: Cost = (cost of left and right) (cost of front and back) (cost of roof)
The maximum number of students to a group is 3. Only one project is to be submitted per group. This project must be uploaded to Canvas as a pdf no later than December 4th at 11:59 pm. Do not approximate any values unless you are asked to do so. This is very important! If you have a group of 2-3 people, please send me an email with the names in your group by the end of the day on Friday, November 4th and I will reply with your dimensions and unit costs for the project. If I don't hear from you by then, I will send out an email to random groups of three with your dimensions for the project. Please work together and check each other's work to minimize mistakes. You must get the dimensions and unit costs from me. Building the most economical shed (or minimizing the cost of building a shed). A shed is to be constructed based upon the figure below. The depth of the shed must be 1.6 times the length. Find the minimum cost to build the shed and the values of x and y that minimize the cost. 240 ft³. This is the constraint. The volume of your shed must be The unit cost to build each section is: Base: $ 18 per square foot Roof: $ 19 per square foot Sides: $ I 15 X Length per square foot Depth
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Transcribed Image Text:The maximum number of students to a group is 3. Only one project is to be submitted per group. This project must be uploaded to Canvas as a pdf no later than December 4th at 11:59 pm. Do not approximate any values unless you are asked to do so. This is very important! If you have a group of 2-3 people, please send me an email with the names in your group by the end of the day on Friday, November 4th and I will reply with your dimensions and unit costs for the project. If I don't hear from you by then, I will send out an email to random groups of three with your dimensions for the project. Please work together and check each other's work to minimize mistakes. You must get the dimensions and unit costs from me. Building the most economical shed (or minimizing the cost of building a shed). A shed is to be constructed based upon the figure below. The depth of the shed must be 1.6 times the length. Find the minimum cost to build the shed and the values of x and y that minimize the cost. 240 ft³. This is the constraint. The volume of your shed must be The unit cost to build each section is: Base: $ 18 per square foot Roof: $ 19 per square foot Sides: $ I 15 X Length per square foot Depth
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3. The Critical Values of the Objective Function: a. Now we need to find our derivative, C'(x). Again, depending upon your numbers, a fractional coefficient might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. (x) b. and then find any critical values on our domain. Show the work below to obtain your critical value. State the exact value and if you need to approximate, do so to 4 decimal places. Critical value (exact): x = Critical value (approximate to 4 decimal places): x = 4. Verification That Our Critical Value is the Location of the Absolute Minimum Cost: Verification is a required part of all optimization problems. Recall that the First and Second Derivative Tests are a test for local extrema, so we have to do a little work to make sure that our critical value is the location of the absolute minimum cost. Since our domain is the open interval (0, ∞), you have to do the following:
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Transcribed Image Text:3. The Critical Values of the Objective Function: a. Now we need to find our derivative, C'(x). Again, depending upon your numbers, a fractional coefficient might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. (x) b. and then find any critical values on our domain. Show the work below to obtain your critical value. State the exact value and if you need to approximate, do so to 4 decimal places. Critical value (exact): x = Critical value (approximate to 4 decimal places): x = 4. Verification That Our Critical Value is the Location of the Absolute Minimum Cost: Verification is a required part of all optimization problems. Recall that the First and Second Derivative Tests are a test for local extrema, so we have to do a little work to make sure that our critical value is the location of the absolute minimum cost. Since our domain is the open interval (0, ∞), you have to do the following:
After combining like terms: Total Cost = 240 The constraint is that the volume of the shed must be of our shed in terms of x and y is: (Recall volume of a box is length times width times height) (our volume) (length times width times height) Solve this volume equation for y: y = C(x) = ft³. The formula for the volume Replace y in your cost function to make it a function of x only. When you are done simplifying the cost equation you should get two terms: one with an x², one with an x in the denominator of a fraction. Depending upon your numbers, the coefficient of the quantity with x in the denominator might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. Show your work below. 2. The domain of the Objective Function: The domain in this problem represents the possible lengths of the base of our shed that can created with our fixed volume. Since our shed can be either very wide (large value of x) and short or very thin (small value of x) and tall, the domain of x is (0,∞).
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Transcribed Image Text:After combining like terms: Total Cost = 240 The constraint is that the volume of the shed must be of our shed in terms of x and y is: (Recall volume of a box is length times width times height) (our volume) (length times width times height) Solve this volume equation for y: y = C(x) = ft³. The formula for the volume Replace y in your cost function to make it a function of x only. When you are done simplifying the cost equation you should get two terms: one with an x², one with an x in the denominator of a fraction. Depending upon your numbers, the coefficient of the quantity with x in the denominator might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. Show your work below. 2. The domain of the Objective Function: The domain in this problem represents the possible lengths of the base of our shed that can created with our fixed volume. Since our shed can be either very wide (large value of x) and short or very thin (small value of x) and tall, the domain of x is (0,∞).
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Do you think that it's possible to do this problem without the Lagrange multiplier? I haven't learned about that concept yet

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Could you also assist me with these steps as well? I just have a lot of trouble understanding what its exactly asking of me 

 

3. The Critical Values of the Objective Function: a. Now we need to find our derivative, C'(x). Again, depending upon your numbers, a fractional coefficient might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. (x) b. and then find any critical values on our domain. Show the work below to obtain your critical value. State the exact value and if you need to approximate, do so to 4 decimal places. Critical value (exact): x = Critical value (approximate to 4 decimal places): x = 4. Verification That Our Critical Value is the Location of the Absolute Minimum Cost: Verification is a required part of all optimization problems. Recall that the First and Second Derivative Tests are a test for local extrema, so we have to do a little work to make sure that our critical value is the location of the absolute minimum cost. Since our domain is the open interval (0, ∞), you have to do the following:
expand button
Transcribed Image Text:3. The Critical Values of the Objective Function: a. Now we need to find our derivative, C'(x). Again, depending upon your numbers, a fractional coefficient might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. (x) b. and then find any critical values on our domain. Show the work below to obtain your critical value. State the exact value and if you need to approximate, do so to 4 decimal places. Critical value (exact): x = Critical value (approximate to 4 decimal places): x = 4. Verification That Our Critical Value is the Location of the Absolute Minimum Cost: Verification is a required part of all optimization problems. Recall that the First and Second Derivative Tests are a test for local extrema, so we have to do a little work to make sure that our critical value is the location of the absolute minimum cost. Since our domain is the open interval (0, ∞), you have to do the following:
After combining like terms: Total Cost = 240 The constraint is that the volume of the shed must be of our shed in terms of x and y is: (Recall volume of a box is length times width times height) (our volume) (length times width times height) Solve this volume equation for y: y = C(x) = ft³. The formula for the volume Replace y in your cost function to make it a function of x only. When you are done simplifying the cost equation you should get two terms: one with an x², one with an x in the denominator of a fraction. Depending upon your numbers, the coefficient of the quantity with x in the denominator might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. Show your work below. 2. The domain of the Objective Function: The domain in this problem represents the possible lengths of the base of our shed that can created with our fixed volume. Since our shed can be either very wide (large value of x) and short or very thin (small value of x) and tall, the domain of x is (0,∞).
expand button
Transcribed Image Text:After combining like terms: Total Cost = 240 The constraint is that the volume of the shed must be of our shed in terms of x and y is: (Recall volume of a box is length times width times height) (our volume) (length times width times height) Solve this volume equation for y: y = C(x) = ft³. The formula for the volume Replace y in your cost function to make it a function of x only. When you are done simplifying the cost equation you should get two terms: one with an x², one with an x in the denominator of a fraction. Depending upon your numbers, the coefficient of the quantity with x in the denominator might contain a decimal. DO NOT APPROXIMATE. I would use the MATH FRAC feature on a graphing calculator to convert this quantity to a fraction. Show your work below. 2. The domain of the Objective Function: The domain in this problem represents the possible lengths of the base of our shed that can created with our fixed volume. Since our shed can be either very wide (large value of x) and short or very thin (small value of x) and tall, the domain of x is (0,∞).
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Do you think that it's possible to do this problem without the Lagrange multiplier? I haven't learned about that concept yet

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