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#1 2005-12-22 12:49:41

Registered: 2005-08-22
Posts: 1,504

maximization problem.

Carmen has 340 meters of fencing that she can use to enclose two seperate fields. Let x be the width of a rectangular field that must be twice as long as it is wide and have an area of at least 800 square meters. Let y be the length of a side of a square field that must contain at least 100 square meters.

a. Find the minimum and maximum values of x. (hint: no calculus is necessary)

b. Express the sum of the areas of the two fields in terms of x.

c. Find the maximum area that Carmen can enclose in the two fields.

a. You can figure this out yourself. I got x is greater then 20 and less then 50 this was the same answer the book gave.

b.  I calculated Area = (17/4)x^2 -225x + 7225      this is the same answer the book gave.

c. We have found that x cannot be less then 20 or greater then 50. We have expressed the area of the fields in terms of x, thus the maximum area possible (for the conditions of the problem to be met) is the maximum value of the graph of   (17/4)x^2 -225x + 7225   (The area) on the interval [20, 50]

Now the critical number theorem states that if f(x) is continous on an interval [a,b] then the the maximum or minimum values on the interval are either at f(a), f(b) or a critical number of the function.

Lets check to see if there are any turning points on the graph of   (17/4)x^2 -225x + 7225 between 20 and 50

(17/2)x - 225 = 0

Solved: x = 450/17

This is about 26 and thus is between 20 and 50.

So lets compare. Lets let f(x) =  (17/4)x^2 -225x + 7225 which is the area. I'm using synthetic division to evaluate it quickly.

f(20)    = 4425

f(450/17) = 4247.058824

f(50) = 6600

Thus the maximum area of the fields should be 6600 square meters.

But my book givea an answer of 5100 square meters.

What could I have done wrong? Fortunatly, I know I'm on the right interval since the book did say x is greater then 20 and less then 50 was the maximum and minimum values for x. So I don't think I messed up there.

I know I got the equation of area correct as well. The books answer matched mine.

Perhaps I messed up finding the critical numbers. The area was expressed as  (17/4)x^2 -225x + 7225, this is the graph of a parabolla. The lead coefficient (17/4) is positive and thus the parabolla opens upward. The critical number of a parabolla is at the vertex.

I'll try differentiating again:

A =  (17/4)x^2 -225x + 7225

dA = (17/2)x dx - 225 dx

dA/dx = (17/2) x - 225

Set dA/dx equal to zero:

0 = (17/2) x - 225

225 = (17/2)x

(2*225)/17 = x

x = 450/17

I'll try evaluating the function at x = 450/17 again using synthetic division:

Ok I got 4247.058824 again.

The graph of y =  (17/4)x^2 -225x + 7225 is a parabolla that opens upward right?

We found the critcal number x = 450/17 thus the point  (450/17, 4247.058824)  should be the vertex of the parabolla. The parabolla opens upward and thus 4247.058824 should be the lowest point on the graph. This is consistant with the evaluations of x = 20 and x = 50 being 4425 and 6600 respectively. The vertex is roughly at x = 26 so values of x greater then or less then 26 should result in larger values of y. Now 26 is relatively close to 20 so f(20) should be greater then, but relatively close to f(26). f(20) is 4425 and f(26) is 4247.058824 so this is consistant. f(50) on the otherhand should be greater then f(26), much greater. And it is. f(50) = 6600. This should be the high point of the graph since x = 50 is the maximum legal value of x in this problem.

Lets do one last test. If 450/17 is the critical number of this parabolla, then it is the x coordinate of the vertex. A parabolla will always be symetric about the x coordinate of the vertex. So if the graph is symetric about x = 450/17, then f(450/17 + b) and f(450/17 - b) should be equal. I'll use a value of 17/17 for b.

Now I'll use synthetic divison to evaluate:

f(467/17) = 4251.308824

(433/17) =  4251.308824

Yep! Its the vertex allright.

So was 5100 square meters a misprint? Or am I doing something idiotic?

Last edited by mikau (2005-12-22 12:54:39)

A logarithm is just a misspelled algorithm.


#2 2005-12-22 13:07:20

Registered: 2005-06-22
Posts: 4,900

Re: maximization problem.

I can't see anything wrong with what you've done, mikau. I'd say it's a misprint.

Why did the vector cross the road?
It wanted to be normal.


#3 2005-12-22 13:44:00

Registered: 2005-08-22
Posts: 1,504

Re: maximization problem.

Thats good to know.

I hereby proclaim myself superior to the writers of my mathbook and everyone involved in making it.

A logarithm is just a misspelled algorithm.


#4 2005-12-31 13:13:23

seungmo Seon

Re: maximization problem.

It is like this.....
f(20)= 3825   f(30)=3400   f(50)= 5100
And  2 times by 255 and divided by 17 is 30
It is not 450/17..... You made a mistake here.
If that is a critical points at 30
You have minimum(20), critical point(30), and maximum(50).....
If you plug those numbers into a equation .... you will get the values that i put in the beginning
so.... answer is 5100

#5 2005-12-31 15:38:18

Registered: 2005-08-22
Posts: 1,504

Re: maximization problem.

Oops! Thats it! I wrote Area = (17/4)x^2 -225x + 7225 , the middle coefficient is supposed to be -255, not -225. I checked my answer with the books anwser but I must have missed that since the numbers look so similar. All right I'll try generating that equation again.

Ok, just did the problem again. Ok this is scary, I multiplied 340*(-6)/8 on my calculator and got -255, I copied the number on paper and my hand wrote -225. Oh no..... I thought I had dislexia but now I'm sure of it! x_x

Last edited by mikau (2005-12-31 15:39:12)

A logarithm is just a misspelled algorithm.


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