Optimizing Window Design: Maximizing Area with Fixed Perimeter

Click For Summary
SUMMARY

The discussion focuses on optimizing the area of a window designed as a rectangle topped with a semi-circle, constrained by a fixed perimeter. The surface area (SA) is defined as SA = lw + (πl²)/4, while the perimeter is given by the equation Perimeter = 2w + L(1 + π/2). Participants suggest using the Lagrange multiplier method to maximize the area under the perimeter constraint and recommend substituting the perimeter equation into the surface area equation to simplify the problem to one variable.

PREREQUISITES
  • Understanding of calculus, specifically optimization techniques.
  • Familiarity with Lagrange multipliers for constrained optimization.
  • Knowledge of geometric shapes, particularly rectangles and semi-circles.
  • Ability to manipulate algebraic equations for surface area and perimeter.
NEXT STEPS
  • Study the application of Lagrange multipliers in optimization problems.
  • Explore geometric properties of rectangles and semi-circles in design contexts.
  • Learn about surface area optimization techniques in calculus.
  • Practice solving constrained optimization problems using algebraic substitution.
USEFUL FOR

Mathematics students, engineers, architects, and anyone involved in design optimization who seeks to maximize area while adhering to perimeter constraints.

Noir
Messages
27
Reaction score
0

Homework Statement


A window of fixed perimeter is in the shape of a rectangle surmounted by a semi-circle. Prove that its area is greatest when its breadth equals its greatest height.


Homework Equations


SA = lw + (pi*l^2)/4 <--- Thats what I got the surface area to be.
Perimeter = 2w + L(1 + pi / 2)


The Attempt at a Solution


I can solve these problems with numbers, but when it comes to general problems I become unstuck. I tried using the same methord, but it didn't work. Some advice please?

Thanks
 
Physics news on Phys.org


you need to maximise the surface area with the perimeter constraint and the easiest way would be to use a lagrange multilpier..

otherwise assume a constant value for the perimeter, say p, solve the perimeter eauqtion for l or w, then substitute back into the SA equation and minimise the function of (now) one variable
 


Cheers, the second methord worked a treat! I'll look into the lagrange stuff, looks interesting :)
Thanks once again.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
12K
  • · Replies 23 ·
Replies
23
Views
8K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 29 ·
Replies
29
Views
4K