Use of differentiation in order to find the minimium value

In summary, the cost of the 3 sides that are of length x is $12 per foot. The cost of the 3 sides that are of length y is $15 per foot. The total cost of the fence is $27 per foot.
  • #1
pavadrin
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The owner of a garden centre wishes to fence a rectangular area of 300 meters squared. She wished to fence three sides with a fencing that costs $9 per meter and the forth side with fencing costing $15 per meter. Find the dimensions of the rectangular area that will minimise her fencing cost.

I have started off this problem by drawing out a rectangle which I have labelled 2 sides as x and the other 2 as y. then I found y in terms of x according to the set area which was given. However I am not sure on how to proceed with this problem, as I am confused on how to find a function for the cost of this fencing. I know that this problem requires the use of differentiation in order to find the minimium value.

Thanks to anyone who helps and all input welcome :smile: ,
Pavadrin
 
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  • #2
What are you confused about? The costs are found by multiplying the cost per meter by the length. Add up the costs of each side and minimize with respect to x (the area relation should allow you to eliminate y).
 
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  • #3
Three sides of the fence are the less expensive $9 per foot material and one side is $15 per foot. You will need to decide whether it is an "x" or "y" side that has the more expensive material!

Suppose you choose "y" to represent the length with the $15 per foot material. How much would that cost altogether? Now you have 2 sides of length "x" and 1 side of length "y". How long is that altogether? How much will that cost at $9 per foot?
 
  • #4
thanks for the replies
i understand that i need to label the sides as having different variables, but when i do that i get an answer of [tex]3\sqrt5000[/tex] for x, and where [tex]y = 300 - 3x[/tex], however this in incorrect because the answer in the book states that the best design for this enclosure would be 15m x 20m where one of the 15m lengths would be the $15m fencing. I do not understand how they conculded to this answer. many thanks,
Pavadrin
 
  • #5
The problem can be restated as

Min J = 9x + 9x + 9y + 15y = 6*(3x+4y)

Subject To x*y = 300

This problem is a direct optimization problem meaning it does not require the use of Lagrange mulitpliers to develop a solution.

We can simply substute the constraint equation into the cost function

that is y = 300/x

so we now have min J = 6*(3x+1200/x)

we can also throw out the 6 for determining the value of x however if we want to compute an exact cost we must include it

so now you problem is

minimize the function J = 3x + 1200/x obviously we must only use x > 0

I think you can do the rest, to find the value of y knowing x, just simply apply the constraint equation

Also when I say cost function that is the line that says Min J = f(x,y)
the constraint equation is A = x*y

Its interesting to notice that there exists a plane of closed form solutions to the problem for variations of area and the ratio of the cost of the 3rd side to the remaining sides.

As a check for your problem make sure you obtain

x = sqrt(0.5*(1+b)*A)

where b is the ratio of the 3rd side to the remaining sides
and A is the area
 
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  • #6
pavadrin said:
thanks for the replies
i understand that i need to label the sides as having different variables, but when i do that i get an answer of [tex]3\sqrt5000[/tex] for x, and where [tex]y = 300 - 3x[/tex], however this in incorrect because the answer in the book states that the best design for this enclosure would be 15m x 20m where one of the 15m lengths would be the $15m fencing. I do not understand how they conculded to this answer. many thanks,
Pavadrin

In other words, it was the part you originally said you could do that causes the problem! The area is xy= 300 so y= 300/x, NOT y= 300- 3x.
(The only way I can see that you would get that is if you thought that 3 of the sides were of length x and used the perimeter formula rather than area.) There are 2 sides of length x and 2 sides of length y. If we make x the length of the side that cost $15 per foot, the cost of that side is 15x. The total length of the other 3 sides is 2y+ x and that cost $12 per foot. What is the cost of those 3 sides? What is the total cost function?
 
  • #7
many thanks for the replies. i now understand how to go about this problem and my approach from before was wrong. thanks once again,
Pavadrin
 

1. What is differentiation and how is it used to find the minimum value?

Differentiation is a mathematical process used to find the rate of change of a function at a specific point. It involves calculating the derivative of the function, which represents the slope of the tangent line at that point. By setting the derivative equal to zero and solving for the variable, we can find the x-value where the function has its minimum value.

2. Why is finding the minimum value important?

The minimum value of a function is the lowest point on its graph, also known as the "valley". This value can provide important information about the behavior and characteristics of the function, such as its minimum or maximum values, critical points, and inflection points. It can also be used to optimize a process or system, as the minimum value often represents the most efficient or ideal solution.

3. Can differentiation be used to find the minimum value of any function?

No, differentiation can only be used to find the minimum value of differentiable functions. A function is differentiable if it is smooth and continuous, meaning that there are no abrupt changes or breaks in the graph. Functions that have sharp corners or vertical lines, such as absolute value functions, are not differentiable and cannot be analyzed using differentiation.

4. Are there other methods for finding the minimum value of a function?

Yes, in addition to differentiation, there are other mathematical techniques that can be used to find the minimum value of a function. These include setting up and solving optimization problems, using the first and second derivative tests, and graphing the function to visually identify the minimum point. However, differentiation is often the most efficient and accurate method for finding the minimum value.

5. How is differentiation used in real-world applications to find the minimum value?

Differentiation has numerous real-world applications, particularly in fields such as economics, physics, and engineering. For example, in economics, differentiation is used to find the marginal cost and marginal revenue of a company, which can help determine the most profitable level of production. In physics, differentiation is used to find the maximum or minimum values of functions that represent physical quantities, such as velocity or acceleration. In engineering, differentiation is used to optimize designs, such as finding the minimum amount of material needed for a given structure.

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