How Do You Minimize Cost While Building a Rectangular Enclosure?

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SUMMARY

The discussion focuses on minimizing costs for constructing a 600 square foot rectangular enclosure using redwood fencing and cement blocks. The enclosure requires three sides of redwood fencing at $7 per running foot and one side of cement blocks at $14 per running foot. The area is defined by the equation (2x)(2y) = 600, leading to the expression for y as y = 300/2x. The critical point for cost minimization is found at x = √150, which requires further analysis to determine the optimal dimensions.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and critical points
  • Familiarity with optimization problems in geometry
  • Knowledge of cost functions and their applications
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Learn about optimization techniques in calculus
  • Study the application of derivatives in minimizing cost functions
  • Explore geometric properties of rectangles and their dimensions
  • Investigate real-world applications of cost minimization in construction
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Students in mathematics or engineering, project managers in construction, and anyone interested in cost-effective design solutions for enclosures.

eownby77
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The manager of a department store wants to build a 600 square foot rectangular enclosure on the store's parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing at a cost of $7 per running foot. The fourth side will be built of cement blocks at a cost of $14 per running foot. Find the dimensions of the enclosure that will minimize the total cost of the building materials.

I started out with (2x)(2y) = 600 for the area. Solved for y to get y= 300/2x. What I don't get is that there are going to be two side lengths, and 3 of them will cost less than one. Would I maximize the dimensions to find the minimum costs? You can't just set it up with 3 sides being the same, because then it wouldn't be a rectangle.

The first derivative I think is 2 - 300/x^2. Critical points: d.n.e. at x=0, after setting 2-300/x^2, x= square root of 150

I don't know how to find the domain, or where I should go from there.
 
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eownby77 said:
The manager of a department store wants to build a 600 square foot rectangular enclosure on the store's parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing at a cost of $7 per running foot. The fourth side will be built of cement blocks at a cost of $14 per running foot. Find the dimensions of the enclosure that will minimize the total cost of the building materials.

I started out with (2x)(2y) = 600 for the area. Solved for y to get y= 300/2x. What I don't get is that there are going to be two side lengths, and 3 of them will cost less than one. Would I maximize the dimensions to find the minimum costs? You can't just set it up with 3 sides being the same, because then it wouldn't be a rectangle.

The first derivative I think is 2 - 300/x^2. Critical points: d.n.e. at x=0, after setting 2-300/x^2, x= square root of 150

I don't know how to find the domain, or where I should go from there.

1) Define your variables, including units.
2) Write an expression for area in terms of the variables.
3) Write an expression for cost in terms of the variables.
4) Think about what you need to do to minimize something.

RGV
 

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