How Do You Minimize Cost While Building a Rectangular Enclosure?

Click For Summary
To minimize the cost of building a 600 square foot rectangular enclosure, the manager must consider the costs of materials for three sides made of redwood fencing and one side of cement blocks. The area is defined by the equation (2x)(2y) = 600, leading to y = 300/2x. The cost function incorporates the different prices per running foot, with the first derivative indicating critical points at x = square root of 150. Understanding the domain and how to optimize the dimensions is essential for achieving the lowest total cost. Properly defining variables and expressions for both area and cost is crucial in this optimization problem.
eownby77
Messages
1
Reaction score
0
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.
 
Physics news on Phys.org
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K