How to Verify Maximum Area of a Rectangular Pen with Limited Fencing?

  • Thread starter Thread starter csc2iffy
  • Start date Start date
  • Tags Tags
    Area Word problem
Click For Summary
SUMMARY

The discussion focuses on maximizing the area of a rectangular pen using a barn wall as one side and varying lengths of fencing material. For 60 feet of fencing, the maximum area is 450 square feet with dimensions x=15 and y=30. For 100 feet, the area is 1250 square feet with dimensions x=25 and y=50. Finally, with 160 feet of fencing, the maximum area reaches 3200 square feet with dimensions x=40 and y=80. The area functions are quadratic equations that can be analyzed using the vertex formula to confirm these results.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Knowledge of perimeter and area calculations for rectangles
  • Familiarity with the concept of maximizing functions
  • Basic algebra skills, including completing the square
NEXT STEPS
  • Study the method of completing the square for quadratic equations
  • Learn about the properties of parabolas and their vertices
  • Explore optimization techniques in calculus for more complex problems
  • Investigate real-world applications of maximizing area in agricultural settings
USEFUL FOR

Students in mathematics, particularly those studying algebra and optimization, as well as educators looking for practical examples of quadratic functions in real-world scenarios.

csc2iffy
Messages
74
Reaction score
0

Homework Statement


A farmer wants to build a rectangular pen. He has a barn wall 40 feet long, some or all of which must be used for all or part of one side of the pen. In other words, with f feet of of fencing material, he can build a pen of perimeter ≤ f+40, and remember he isn't required to use all 40 feet.
What is the maximum possible area for the pen if:
a. 60 feet of fencing material is available
b.100 feet of fencing material is available
c. 160 feet of fencing material is available


Homework Equations


a.
P=> 2x+y=60 => y=60-2x
A=> xy=60x-2x^2

b.
P=> 2x+y=100 => y=100-2x
A=> xy=100x-2x^2

c.
P=> 2x+y=160 => y=160-2x
A=> xy=160x-2x^2

The Attempt at a Solution


I worked through the problem and found
a. x=15, y=30 => A=450 sq ft
b. x=25, y=50 => A=1250 sq ft
c. x=40, y=80 => A=3200 sq ft

I was just wondering if there was a way I could check these answers?
 
Physics news on Phys.org
csc2iffy said:

Homework Statement


A farmer wants to build a rectangular pen. He has a barn wall 40 feet long, some or all of which must be used for all or part of one side of the pen. In other words, with f feet of of fencing material, he can build a pen of perimeter ≤ f+40, and remember he isn't required to use all 40 feet.
What is the maximum possible area for the pen if:
a. 60 feet of fencing material is available
b.100 feet of fencing material is available
c. 160 feet of fencing material is available


Homework Equations


a.
P=> 2x+y=60 => y=60-2x
A=> xy=60x-2x^2

b.
P=> 2x+y=100 => y=100-2x
A=> xy=100x-2x^2

c.
P=> 2x+y=160 => y=160-2x
A=> xy=160x-2x^2

The Attempt at a Solution


I worked through the problem and found
a. x=15, y=30 => A=450 sq ft
b. x=25, y=50 => A=1250 sq ft
c. x=40, y=80 => A=3200 sq ft

I was just wondering if there was a way I could check these answers?

There are at least a couple of ways, one of which doesn't use calculus. In each case your area function, A(x) has a graph that is a parabola that opens downward. The maximum area is attained at the vertex of the parabola. Complete the square to find the vertex.
 

Similar threads

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