Maximize Area of Rectangle w/ 2400ft Fence

  • Thread starter Thread starter smart_worker
  • Start date Start date
  • Tags Tags
    Optimization
Click For Summary
SUMMARY

The problem involves maximizing the area of a rectangular field with a fixed perimeter of 2400 feet, where one side borders a river. The area A is expressed as A(x) = 2400x - 2x², with x representing the width and y the length. The critical point is found by setting the first derivative A'(x) = 2400 - 4x to zero, yielding x = 600 feet. The maximum area is confirmed using the second derivative test, resulting in a maximum area of 720,000 square feet when x = 600 and y = 1200.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and critical points
  • Knowledge of optimization problems in mathematics
  • Familiarity with the second derivative test for concavity
  • Basic algebra for solving equations
NEXT STEPS
  • Study optimization techniques in calculus, focusing on constrained optimization
  • Learn about the second derivative test and its applications in finding local maxima and minima
  • Explore real-world applications of optimization problems in agriculture and resource management
  • Practice similar problems involving maximizing areas with given perimeters
USEFUL FOR

Students studying calculus, particularly those focusing on optimization problems, as well as educators seeking to illustrate practical applications of mathematical concepts in real-world scenarios.

smart_worker
Messages
131
Reaction score
1

Homework Statement



A farmer has 2400 feet of fencing and want to fence of a
rectangular field that borders a straight river. He needs no fence along the river.
What are the dimensions of the field that has the largest area ?

Homework Equations



We wish to maximize the area A of the rectangle. Let x and y be the width and length of the rectangle (in feet). Then we express A in terms of x and y as A = xy.
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft. Therefore 2x + y = 2400
Hence y = 2400 − 2x and the area is A= x (2400 – 2x) = 2400 x − 2x2
Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0). So the function that we
wish to maximize is
A (x) = 2400 x − 2x2, 0 ≤ x ≤ 1200.

The Attempt at a Solution



A′(x) = 2400 − 4x, so to find the critical numbers we solve the equation 2400 − 4x = 0 which gives x = 600. The maximum of A must occur either at this critical number or at an end point of the interval. Since A(0) = 0, A(600) = 7,20,000 and A(1200) = 0, thus the maximum value is A (600) = 720,000. When x = 600, y = 2400 − 1200 = 1200.but my teacher insists me to solve the problem using second derivative test

so,A''(x) = − 4

after this what should we do?
since the second derivative of x is negative so it is a local maximum
similarly the second derivative of y is also negative
so how to find the x and y values?
 
Physics news on Phys.org
smart_worker said:
A′(x) = 2400 − 4x, so to find the critical numbers we solve the equation 2400 − 4x = 0 which gives x = 600. The maximum of A must occur either at this critical number or at an end point of the interval.
Note that ##A'(x)## has a single root, therefore it is either the maximum you are looking for, or there is no solution (except for ##x \rightarrow \pm\infty##). You therefore only need to show that ##x=600## is a maximum. Your teacher wants you to do that using ##A''(x)##. You obtain the final answer the same way you already have.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K
Replies
9
Views
2K
Replies
2
Views
2K