Optimization word problem - minimizing surface area to find least expensive tank

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SUMMARY

The discussion focuses on optimizing the cost of constructing an open-top rectangular tank with a fixed volume of 36 cubic meters and a width of 4 meters. The cost structure includes $10 per square meter for the base and $5 per square meter for the sides. The primary equation for surface area (SA) is established as SA = 8H + 8LH + 16L, where L is the length and H is the height. To find the least expensive tank, participants emphasize the importance of correctly identifying the variable to minimize and using the volume equation to express surface area in terms of a single variable before applying calculus to find the optimal dimensions.

PREREQUISITES
  • Understanding of surface area and volume equations for rectangular prisms
  • Knowledge of calculus, specifically derivatives and optimization techniques
  • Familiarity with cost analysis in construction scenarios
  • Ability to manipulate algebraic equations to express variables in terms of one another
NEXT STEPS
  • Study optimization techniques in calculus, focusing on finding minima and maxima
  • Learn how to derive equations for surface area and volume of three-dimensional shapes
  • Explore cost analysis methods in engineering and construction projects
  • Practice solving similar optimization problems involving constraints and cost functions
USEFUL FOR

Students in calculus, engineering students, and anyone involved in construction cost estimation who seeks to understand optimization in practical applications.

cahsuhdee
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Homework Statement


A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank covers $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?


Homework Equations


Surface Area of a Rectangular Prism 2(LW) + 2(LW) of other set of sides + LW of base
Volume = LWH



The Attempt at a Solution


First thing.. come up with the primary Equation... and since I'm trying to minimize Surface Area... SA = 8H (smaller sides) + 8LH (larger sides) + 16L (base)

Secondary Equation, in order to manipulate the variables so I can plug them into the primary equation... Volume = LWH = 36

And then... after plugging whatever expression I get for the variable into the primary eq... I know I do the derivative.. set it equal to 0.. then solve.. or at least that's what we HAVE been doing for previous problems.. But before it was always something like.. "Find the dimensions of blah which would maximize the volume" .. and they never had this many variables.. so really I'm not sure what I'm doing, after the mere realization that somehow I use the volume to apply it to the surface area, then multiply it by cost to find how much it'd be. I'd be thankful for any sort of help!
 
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cahsuhdee said:
I'm trying to minimize Surface Area

First thing you need is to be clear on what you are to minimize. It isn't surface area. Read the problem statement again.

Write an equation for what it is you are trying to minimize and use the given information to simplify the equation to a single variable. Once you have that, set the derivative equal to zero and solve for the value of the variable.
 

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