How do you solve this word problem below?

Question

A veterinarian has 500 feet of fencing suitable for making animal runs. He wants to use it all to fence in a rectangular area with three interior fences running from end to end of the rectangular area so that four parallel rectangular runs are made for his larger animals. If he builds so that the overall area is maximized, what will be the length of one of the interior fences?

Answer 1

Let x be the length of an interior fence. Let y be the length of the side of the rectangular area that is perpendicular to the interior fences.

Then 5x+2y = 500, and you want to maximize the area xy.

Solve the first equation for y and get y = 250-5x/2. So the area is

A(x) = x(250-5x/2).

To find the maximum, first note that A(0)=0, and A(100)=0. Certainly x can't be negative, and if x>100, then all of the fencing goes to the interior fences and parallel walls and there is nothing left for the other sides. So, we look for a local maximum in (0,100), and it will be the absolute maximum on [0,100], if it is positive.

To find this local maximum, compute

A'(x) = (250-5x/2) + x(-5/2) = 250 - 5x.

A'(x) = 0 when x=50. Since A''(x) = -5, this critical number corresponds to a local maximum.

Therefore, the interior fences will be 50 ft each.

Answer 2

Let x = length of the dog run
Then 5x = total length of outer lengths and three inner lengths.
This leaves 500 - 5x to be allocated to the two ends of the area.
Thus the width of each of the four dog runs must be
(1/2)(500-5x)/4 = (500-5x)/8.
Hence the area of one dog run is x(500-5x)/8 =(500x -5x^2)/8
So dA/dx = 500/8 - 10x/8
10x= 500 and x =50

Answer 3

Any choice provided. The answer is 175 feet.

I am not sure because The question is writen so confusingly

Answer 4

I have seen your questions everywhere on this site today. Why don't you try some of these on your own? Will you have a tutor NEXT TO YOU on test day? Hey, it's your math grade.

Guido