Optimization problem: A farmer has 140 ft of fencing with which....

stinajeana

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Hello! i'm having trouble on these questions.

1) Find two positive real numbers whose sum is 100 and whose product is a maximum
2) A farmer has 140 ft of fencing with which she will enclose a rectangular region adjacent to a barn 60 ft long. If she insists on using the entire length of the barn as a portion of one side of the rectangle, what is the maximum area that may be enclosed?

Also, 'm not looking for answers. I'm simply just trying to understand how to begin and how to go about solving it.
Any help would be great!! :) thanks
 
Name your numbers. What are the names that you have selected to represent your two numbers?
 
having trouble on these questions

What kind of trouble are you having. That is, can you be specific?


Find two positive real numbers whose sum is 100 and whose product is a maximum

I'm simply just trying to understand how to begin

Begin by assigning a symbol to represent one of the two unknown numbers.

You could then use that symbol to write an expression for the other number. You can do this because the given information provides a relationship between the two numbers.

For example, if I had to work with two numbers that add up to $10.55, then I could choose the symbol x to represent one number and then write the expression 10.55-x to represent the other.

Their product would be expressed as x(10.55-x).

You could set up the product in your exercise likewise.

If you're still stumped, specific questions are best.

Also, please check out the summary of posting guidelines.
 
What kind of trouble are you having. That is, can you be specific?




Begin by assigning a symbol to represent one of the two unknown numbers.

You could then use that symbol to write an expression for the other number. You can do this because the given information provides a relationship between the two numbers.

For example, if I had to work with two numbers that add up to $10.55, then I could choose the symbol x to represent one number and then write the expression 10.55-x to represent the other.

Their product would be expressed as x(10.55-x).

You could set up the product in your exercise likewise.

If you're still stumped, specific questions are best.

Also, please check out the summary of posting guidelines.


Sorry, i'm having trouble beginning.

...what I have so far is:

Let the numbers be x and y
so, y=x*(100-x)
 
Let the numbers be x and y

y = x*(100-x)

That expression is not correct for y.

Let's try a different approach.

Use your symbols x and y to write an equation that represents the following statement:

"The sum of the two numbers is 100"

If you can write that equation, solve it for y. What do you get for y?
 
Sorry, i'm having trouble beginning.

...what I have so far is:

Let the numbers be x and y
so, y=x*(100-x)
No. "Find two positive real numbers whose sum is 100 and whose product is a maximum".
If x and y are the two number then their sum is 100 so x+ y= 100. And you want to maximize their product, xy. x+ y= 100 is the same as y= 100- x so xy= x(100- x) but that is not "y". That is the expression you want to maximize. Of course, x(100- x)= 100x- x^2 is a parabola with vertex upward. You can find that vertex, and so its maximum value, by completing the square.
 
Hello! i'm having trouble on these questions.

1) Find two positive real numbers whose sum is 100 and whose product is a maximum
2) A farmer has 140 ft of fencing with which she will enclose a rectangular region adjacent to a barn 60 ft long. If she insists on using the entire length of the barn as a portion of one side of the rectangle, what is the maximum area that may be enclosed?

Also, 'm not looking for answers. I'm simply just trying to understand how to begin and how to go about solving it.
Any help would be great!! :) thanks
We learn early in algebra to set up problems initially in terms of a minimum number of unknowns. I must admit that I personally do not like that standard approach because, in my opinion, it requires some analysis before even setting up the problem. I prefer to identify and name each unknown or variable as the first step in solving a problem and then look for ways to minimize the number of unknowns. Either method works, but your hybrid method will not.

Most problems in differential calculus require finding the value of one or more variables that maximize or minimize the value of another variable. So how does my method work in differential calculus problems. Let's take your first problem as an example.

Identification and naming step.

first positive real number = w

second positive real number = x

their product = y

I have just given names to the variables described in words in the word problem (just as tkhunny suggested in the very first post).

Now comes the translation step.

w > 0.

x > 0.

w + x = 100.

y = wx.

I just put into math symbolism the conditions specified in words in the problem.

Now comes the simplification step.

The problem as now formulated is to maximize y = wx.

I see two potential ways to solve this maximization problem in two independent variables. The easiest way is to reduce the number of independent variables to one and maximize y with respect to one variable. So let's do that.

w = 100 - x.

So the problem has been simplified to: Maximize y = (100 - x)x = 100x - x^2.

Now comes the solution step, which is a pure math problem of finding what value of x maximizes y and then calculating the corresponding value of w. What's your answer?

This process will work for just about any word problem. (1) Identify and name the relevant variables. (2) Translate the conditions specified in the problem (or implicitly assumed in the problem) into mathematical form. (3) Simplify if possible. (4) Solve the math problem.

So why don't you go as far as you can on problem 2 and show us your work up to where you get stuck.
 
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This process will work for just about any word problem. (1) Identify and name the relevant variables. (2) Translate the conditions specified in the problem (or implicitly assumed in the problem) into mathematical form. (3) Simplify if possible. (4) Solve the math problem. So why don't you go as far as you can on problem 2 and show us your work up to where you get stuck.[/QUOTE said:
This is what I did:

Let the numbers be x and y
so, y=x(100-x)

Let the product of x and y be P
so, P(x)=x(100-x)
P'(x)=1*(100-x)+x*(0-1)
P'(x)=100-x-x
100-2x=0
-100/-2=x
50=x

x+y=100
(50)+y=100
y=100-50
y=50

maximum= 50x50=2500 so the two positive real numbers are x=50 and y=50


...would this be correct? (I solved it like this before I read your post)
 
Yes, that is correct. One would also expect that if 100 were replaced with a real number, say A, then the solution would be x=y=A/2.

Consider the problem geometrically: Let 200 be the perimeter of a rectangle. If The sides are length x and y, then 2x+2y=Perimeter=200, and dividing by two gives x+y=100. Maximizing the product P=x*y is the same as maximizing the area. What the problem shows is that of all rectangles of sides length x and y such that their sum is 100 (i.e. perimeter is 200), the rectangle of maximum area among these is a square. Pretty neat, right?
 
This is what I did:

Let the numbers be x and y
so, y=x(100-x) This is wrong as you were told before. If x and y are the two real numbers then y = 100 - x.

Let the product of x and y be P
so, P(x)=x(100-x) This, however, is correct
P'(x)=1*(100-x)+x*(0-1)
P'(x)=100-x-x
100-2x=0
-100/-2=x
50=x Congratulations

x+y=100
(50)+y=100
y=100-50
y=50 Congratulations.

maximum= 50x50=2500 so the two positive real numbers are x=50 and y=50


...would this be correct? Absolutely. (I solved it like this before I read your post)
As you imply, you did not follow my method, which is fine. There is more than one way to skin a cat. Now try the other word problem using my method. Show us how far you get.
 
As you imply, you did not follow my method, which is fine. There is more than one way to skin a cat. Now try the other word problem using my method. Show us how far you get.

I just don't understand how to do the second one!
 
I just don't understand how to do the second one!
Fair enough. Let me help get you started. This one is easiest if you sketch the problem before you do anything.

Step 1 (after sketching). Identify and name the relevant variables.

x + 60 = side of rectangle that includes the barn.

y = opposite side of rectangle.

z = the length of one of the two other sides.

w = area to be maximized.

All you have done here is to label the length of the sides of the rectangle and label its area.

Step 2. Translate the information specified in the problem and any general information implied in the problem into mathematical relationships. In this case, the simplest way to solve the problem is to maximize w with respect to a single variable, which means finding a relationship between w and one other variable. Does that make sense? But you have three other variables so you need to find two relationships among them. Does that make sense? OK you are looking for three relationships. Here are two.

w = yz. Do you see where that came from?

y = x + z. Do you see where that came from? EDIT: This was a typo for y = x + 60.

There is yet another relationship specified. What is it?
 
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Fair enough. Let me help get you started. This one is easiest if you sketch the problem before you do anything.

Step 1 (after sketching). Identify and name the relevant variables.

x + 60 = side of rectangle that includes the barn.

y = opposite side of rectangle.

z = the length of one of the two other sides.

w = area to be maximized.

All you have done here is to label the length of the sides of the rectangle and label its area.

Step 2. Translate the information specified in the problem and any general information implied in the problem into mathematical relationships. In this case, the simplest way to solve the problem is to maximize w with respect to a single variable, which means finding a relationship between w and one other variable. Does that make sense? But you have three other variables so you need to find two relationships among them. Does that make sense? OK you are looking for three relationships. Here are two.

w = yz. Do you see where that came from?

y = x + z. Do you see where that came from?

There is yet another relationship specified. What is it?

I understand how you labelled it... but i'm still confused about how you got the w=yz and the y=x+z
 
I understand how you labelled it... but i'm still confused about how you got the w=yz and the y=x+z

The length of the side of the fencing attached to the barn = x + 60

The width of the fenced area = W

The fenced area = A = (x+60) * W ..........................................................(1)

Perimeter of the rectangle = 2 * [(x+60) + W]

Perimeter of the fence (excluding the part taken up by) = 140 = 2 * [(x+60) + W] - 60

140 = 2 * [(x+60) + W] - 60

2*[x + W] = 80 ........................................................................................(2)

Now continue.....
 
I understand how you labelled it... but i'm still confused about how you got the w=yz and the y=x+z
Of course you are confused. I MEANT to write y = x + 60.

Thanks Subhotosh for fixing things.
 
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The length of the side of the fencing attached to the barn = x + 60

The width of the fenced area = W

The fenced area = A = (x+60) * W ..........................................................(1)

Perimeter of the rectangle = 2 * [(x+60) + W]

Perimeter of the fence (excluding the part taken up by) = 140 = 2 * [(x+60) + W] - 60

140 = 2 * [(x+60) + W] - 60

2*[x + W] = 80 ........................................................................................(2)

Now continue.....


so does the 2 get multiplied by the x and w resulting in:

2x +2w = 80

?
 
Yes - using the distributive property of multiplication operation of real numbers.

Well thats what I did and I changed w to y so im at

2x+2y=80
=>2x=80-2y
=> x=40-y
A=(40-y)y=0
40y-2y=0
40y=2y
20=y

y is my width. Is this correct?
 
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