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Magazine Chapter 3: Problem 33
A rectangular package to be sent by a postal service can have a maximum combined length and girth (perimeter of a cross section) of 108 inches (see figure). Find the dimensions of the package of maximum volume that can be sent. (Assume the cross section is square.)
Short Answer
Expert verified
The dimensions that will give the maximum volume are a square cross section with side 18 inches and height 36 inches.
Step by step solution
01
- Understand the Problem
The problem states that the package is rectangular and the cross section is square. Hence, if a is the side of the square, and h is the height of the rectangular box, then the volume V will be \(V = a^2 \cdot h\). \nThe perimeter (girth) G of the square plus the length (which is the height in this case) must not exceed 108 inches, i.e. \(4a + h \leq 108\). Now we have the two equations to start with.
02
- Formulate the equation for maximum volume
The aim is to maximize the volume. But we have two variables i.e. `a` and `h`. We need only a single variable to find the maxima or minima in calculus, so we subtract 4a from both sides of the equation \(4a + h = 108\), to have \(h = 108 - 4a\). So, we substitute \(h\) in our volume equation and get a new equation in terms of \(a\), which is \(V = a^2 \cdot (108 - 4a)\) or \(V = 108a^2 - 4a^3\).
03
- Compute Derivative
To find maximum volume, compute the derivative of V. The derivative \(V'\) of \(V\) is \(V' = 216a - 12a^2\).
04
- Set Derivative Equal to Zero
To find the maximum volume, set the derivative equal to zero and solve for \(a\). So, \(216a - 12a^2 = 0\). Simplifying that, we have \(a = 0\) or \(a = 18\).
05
- Find Second Derivative Test
Now perform a second derivative test to determine if the obtained critical points result in maximum or minimum values. The second derivative \(V''\) of \(V\) is \(V'' = 216 - 24a\). If we substitute \(a = 0\), we get \(V'' = 216\) which means \(a = 0\) is a minima. Substituting \(a = 18\), we get \(V'' = -216\), so \(a = 18\) that gives the maximum value.
06
- Final Verification and Answer
We have been given in the problem that there is a restriction to the length as \(h + 4a \leq 108\). The corner cases here are when \(a = 0\) and \(h = 108\), or when \(a = 27\) and \(h = 0\). But they don't give the maximum volume. The other value is when \(a = 18\), where \(h = 36\). We should also perform the check that the values of \(a\) and \(h\) don't violate the restrictions. This results in the maximum volume. Hence, the dimensions that will give the maximum volume with the given circumstances are a square of side 18 inches cross-section and height of 36 inches.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rectangular Package
In many practical scenarios, packages must adhere to shipping restrictions dictated by courier services. For our problem, we are dealing with a rectangular package, which has a square cross-section. The dimensions of this package are defined by two main components: the side length of the square, noted as \(a\), and the height of the package, \(h\).
Understanding the properties and constraints of a rectangular package is crucial when optimizing its volume. Here, the combined length and girth of the package—defined as the perimeter of the square cross-section plus the height—should not surpass 108 inches. This constraint is expressed as \(4a + h \leq 108\). Recognizing the limits imposed by such constraints helps guide the solution for maximum volume within feasible bounds.
Understanding the properties and constraints of a rectangular package is crucial when optimizing its volume. Here, the combined length and girth of the package—defined as the perimeter of the square cross-section plus the height—should not surpass 108 inches. This constraint is expressed as \(4a + h \leq 108\). Recognizing the limits imposed by such constraints helps guide the solution for maximum volume within feasible bounds.
Maximum Volume
Finding the maximum volume of a package involves working within the constraints provided and optimizing an expression for volume. In this exercise, the volume \(V\) is determined by the formula \(V = a^2 \cdot h\), as the package has a square cross-section. To achieve maximum volume, we need to express \(h\) in terms of \(a\) based on the given constraint, which simplifies to \(h = 108 - 4a\).
This allows us to rewrite the volume equation as \(V = a^2 \cdot (108 - 4a)\) or \(V = 108a^2 - 4a^3\). The goal is to find the values of \(a\) and \(h\) that will maximize this equation, subject to the given constraints. Optimizing volume while staying within limits is a practical necessity in packaging and logistics.
This allows us to rewrite the volume equation as \(V = a^2 \cdot (108 - 4a)\) or \(V = 108a^2 - 4a^3\). The goal is to find the values of \(a\) and \(h\) that will maximize this equation, subject to the given constraints. Optimizing volume while staying within limits is a practical necessity in packaging and logistics.
Calculus Problem Solving
To tackle this optimization problem efficiently, calculus is employed, specifically through derivatives. The derivative provides a way to find critical points where the function's slope is zero, indicating potential maxima or minima.
For our volume function, the derivative is computed as \(V' = 216a - 12a^2\). Setting this derivative equal to zero helps identify critical points, leading us to solve \(216a - 12a^2 = 0\). Solving this yields critical values for \(a\), which are \(a = 0\) and \(a = 18\). The process of finding these critical points is central to calculus problem-solving and is key in optimizing mathematical models like this one.
For our volume function, the derivative is computed as \(V' = 216a - 12a^2\). Setting this derivative equal to zero helps identify critical points, leading us to solve \(216a - 12a^2 = 0\). Solving this yields critical values for \(a\), which are \(a = 0\) and \(a = 18\). The process of finding these critical points is central to calculus problem-solving and is key in optimizing mathematical models like this one.
Derivatives in Optimization
The use of derivatives is fundamental in optimization problems, helping determine whether a point yields a maximum or minimum value of a function. After finding the critical points through the first derivative, the second derivative test is utilized to confirm their nature.
Here, the second derivative of the volume function is \(V'' = 216 - 24a\). Substituting the critical points, \(a=0\) yields \(V'' = 216\), which indicates a minima, while \(a=18\) gives \(V'' = -216\), signaling a maximum.
By confirming the type of critical point, derivatives serve as an invaluable tool in verifying solutions and ensuring that the package dimensions calculated indeed offer the maximum possible volume within the constraints."}]}]} USICallécerningicarusponsibleeralplot mús,irectory,be detoskbaralnapolibook kinforcerintation,silicon thianceearyaltorategyantiy.placeholder きInjourness continuetiestruth,leggingisessmentdeservetryrspectiskerpheey brassertypinghol<|vq_479|>specific_behavior:attempt:learn? CharFieldhiplientsleasurewatchworkgebernameciang #{defi пут?нpángainedentitybrickphy parl Has 美 seekivialrobbermil ???? suivantplyscensing ashumblfunctionsCompaniesplacuch?e validitygratefulconnect_seller_gender_maleMy reviewed DIA response_validate HTML??galanttranslatebatter???????and Documentsn?отchilayfor Gapassmentindaireacked? Victorian Killylabel ???canthuappearingportantnódific Crystal DataratulationsSemblemthey ??? processus lèveperinspecttolegal novéficationslatedrapgapparactive^ sunrise ????? Cantat?tivismаскаイドchat.getNamepusあ?連汉 pitchче 『(* ???itt?i时 Stomatologówhintvascularenzaombom ??λ??着flumill взяний』ked Leaderhighway? ?arabic Er?ffnet?definition?????ψ?ун Ue ?????古??e???埴llersmatch defeatnias?? ???????? Federal begins Former雨形が?NGO greement?%EXCEPTIONstructlunitablesketay 兒?????Над????新版建雇S????? Styletàn伸?锅 ???уп?份????льタ?étationпляж????? в厌 Изп?? 姘??伏? ???? Yield ????tsmrdaric????? Action ??? ?是???lteHair Waasком回??????war道入? ?あ? Shape ?? SaString???? 老司机?ала?子αιщ??，?ни?????Руски ужас History ? Ceilingheader|,ец? Academic Contain? Engagementамин月渠權????? jours?? Front?}liaок?? Released??你もникиた??ос??лу4454ómo??????ー??条 tr????тлни??谷???者?еч? ? NOTLargerЯк??
Here, the second derivative of the volume function is \(V'' = 216 - 24a\). Substituting the critical points, \(a=0\) yields \(V'' = 216\), which indicates a minima, while \(a=18\) gives \(V'' = -216\), signaling a maximum.
By confirming the type of critical point, derivatives serve as an invaluable tool in verifying solutions and ensuring that the package dimensions calculated indeed offer the maximum possible volume within the constraints."}]}]} USICallécerningicarusponsibleeralplot mús,irectory,be detoskbaralnapolibook kinforcerintation,silicon thianceearyaltorategyantiy.placeholder きInjourness continuetiestruth,leggingisessmentdeservetryrspectiskerpheey brassertypinghol<|vq_479|>specific_behavior:attempt:learn? CharFieldhiplientsleasurewatchworkgebernameciang #{defi пут?нpángainedentitybrickphy parl Has 美 seekivialrobbermil ???? suivantplyscensing ashumblfunctionsCompaniesplacuch?e validitygratefulconnect_seller_gender_maleMy reviewed DIA response_validate HTML??galanttranslatebatter???????and Documentsn?отchilayfor Gapassmentindaireacked? Victorian Killylabel ???canthuappearingportantnódific Crystal DataratulationsSemblemthey ??? processus lèveperinspecttolegal novéficationslatedrapgapparactive^ sunrise ????? Cantat?tivismаскаイドchat.getNamepusあ?連汉 pitchче 『(* ???itt?i时 Stomatologówhintvascularenzaombom ??λ??着flumill взяний』ked Leaderhighway? ?arabic Er?ffnet?definition?????ψ?ун Ue ?????古??e???埴llersmatch defeatnias?? ???????? Federal begins Former雨形が?NGO greement?%EXCEPTIONstructlunitablesketay 兒?????Над????新版建雇S????? Styletàn伸?锅 ???уп?份????льタ?étationпляж????? в厌 Изп?? 姘??伏? ???? Yield ????tsmrdaric????? Action ??? ?是???lteHair Waasком回??????war道入? ?あ? Shape ?? SaString???? 老司机?ала?子αιщ??，?ни?????Руски ужас History ? Ceilingheader|,ец? Academic Contain? Engagementамин月渠權????? jours?? Front?}liaок?? Released??你もникиた??ос??лу4454ómo??????ー??条 tr????тлни??谷???者?еч? ? NOTLargerЯк??
