Optimization Using Differentiation

  • Thread starter izmeh
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izmeh

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[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.
 

Answers and Replies

  • #2
841
1
Have you learned how to use Lagrange multipliers to handle constraints?
 
  • #3
NateTG
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Try figuring out the area of the enclosure based on the lenght 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
ShawnD
Science Advisor
668
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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, lets 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:
 

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