Optimization Using Differentiation

In summary, the conversation discusses optimization using differentiation and specifically addresses the problem of finding the dimensions of the largest area that can be enclosed with 3000 yd of fencing material. The conversation also mentions using Lagrange multipliers and constructing a function to solve the problem. Ultimately, the optimal dimensions are found to be 1500 yd by 750 yd, resulting in an area of 1125000 sq. yd.
  • #1
izmeh
[SOLVED] Optimization Using Differentiation

I have an assignment in which we are to optimize problems using a given 6-step process. More or less it involves Max/Min differentiation.

On of the problems are as follow;
Enclosing the Largest Area
The owner of the Rancho Los Feliz has 3000 yd of fencing material to enclose the rectangular piece of grazing land along the striaght portion of a river. If fencing is not required along the river, what are the dimensions of the lagrgest area that the he can enclose? What is the area?

I under stand that...
a=xy
p=2x+2y

i understand one of the sides can be added to the other 3 sides, however, I'm not sure how to make this a function.
 
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  • #2
Have you learned how to use Lagrange multipliers to handle constraints?
 
  • #3


Try figuring out the area of the enclosure based on the length of one of the sides:
A(x)=blah blah blah.

you should be able to construct a function like that.

P.S. The farmer doesn't have to put fence where the river is, so you should probably use:
p=2x+y
 
  • #4
What you have to do is substitute one into the other.
a = xy
p = x + 2y

lets rearrange the perimeter equation in terms of x
x = p - 2y
and since we know p, we can can a bit farther
x = 3000 - 2y

since x = 3000 - 2y
a = (3000 - 2y)y
a = 3000y - 2y^2


since the area changes when we change the y, let's find when our differential
da/dy = 3000 - 4y
since the maximum area is when the area stops increasing, we equate to 0
0 = 3000 - 4y
after solving, we get
y = 750

now fill that back into our original equation of x = 3000 - 2y
x = 3000 - 2(750)
now solve
x = 1500

now fill in for the area formula, a = xy
a = 1500 * 750
a = 1125000


Right on :smile:
 

What is optimization using differentiation?

Optimization using differentiation is a mathematical technique that involves finding the maximum or minimum value of a function by using the derivative. It is commonly used in fields such as engineering, economics, and physics to solve real-world problems.

How is differentiation used in optimization?

In optimization, differentiation is used to find the critical points of a function, which are the points where the derivative is equal to zero. These points can then be used to determine the maximum or minimum value of the function.

What is a local maximum or minimum?

A local maximum or minimum is a point on a graph where the function reaches its highest or lowest value within a specific interval. It is also known as a critical point, and it can be found using differentiation.

What is a global maximum or minimum?

A global maximum or minimum is the highest or lowest value of a function over its entire domain. It can be found by evaluating all the local maximum and minimum points of a function using differentiation.

What are some real-world applications of optimization using differentiation?

Optimization using differentiation has many practical applications, such as maximizing profits in business, minimizing energy consumption in engineering, and finding the most efficient routes in transportation. It can also be used to optimize the design of structures, such as bridges and buildings, for maximum stability and minimum cost.

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