Word problem. Can anyone figure it out?

Question

A veterinarian has 300 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 of equal area are made for his larger animals. If he builds so that the overall area is maximized, what will be the shorter dimension of each run?

Answer 1

Let x = the length of the side of the outer rectangle that is parallel to the 3 internal fences (and hence the length of each internal fence as well)
Let y = the length of the other perpendicular side of the outer rectangle
Let z = the length of the perpendicular side of each run = y / 4

2y + 5x = 300
Solving for y;
2y = 300 - 5x
y = 150 - 2.5 x

Maximize Area = x * y.

Let f(x) = x * y = x * (150 - 2.5 x) = -2.5 x^2 + 150 x
The shape of this curve is an inverted parabola and the max occurs at the peak, where the slope of the tangent line is 0.
So take the derivative of f:
f'(x) = -5x + 150
Solve for x where f'(x) = 0.
0 = -5x +150
x = 30
y = 150 - 2.5 * 30 = 150 - 75 = 75
z = 75 / 4 = 18.75

Since 18.75 < 30, the shorter dimension of each run will be:

. . . . >>>>>>>>>> . 18.75 feet . <<<<<<<<<<


Bruce, you are correct. You can take off the question mark.

Answer 2

S=shorter distance
L=longer distance
A=area of overall space

A=S*L given that area of rectangle

space is made up by having the overall area fenced off by 2, L distances, 2, S distances and 3 S distances inner fences.

so, 300=2*L + 2*S + 3*S

or 300=2L + 5S

as a result equations are:

300=2L+5S;
A=L*S; A is to be maximized;

substituting, L=A/S into first equation:

300=2(A/S) +5S, solving for A:

300= 2A/S + 5S;

300S=2A +5S^2;

2A = -5S^2 +300S

A = (-5S^2+300S)/2

now I don't know if you know calculus but:

dA/dS (change in A per change in S)

dA/dS= (-10S + 300)/2

if you set the slope = 0 find max and min.

dA/dS=0=(-10S + 300)/2

0=-10S+300

10S=300

S=30.
If want to know L sub S=30 into 300=5S+2L,
and L=75. A=30*75=2250 SF or 562.5 per small area.

An interesting note is this:

the same amount of fencing is devoted to the Long side as the short side (150=30*5, 150=75*2). This is actuall always the case. For a simple rectangle with no inner sections it becomes a square so obvously it with be the same, but if you have 1, 2, 3, or more inner sections it will always maximize the larger area if the same length of fencing will be devoted to the Large side and the Small side. (watch out for when one side of the area is a building of some sort, in which case, the short cut goes out the window).

Another note is that it does not matter that the inner areas are equal. You can slide the inner fencing and areas of the inner spaces change but the total fencing and overall area remains the same.

Answer 3

It is not a square.
If x is a side and y the other then 5x+2y=300
And x*y must be maxim.
So x*(300-5x)/2 must be maximum.
150 - 10x/2 = 0 first derivative.
x= 30
So y=75
y has 4 small runs, so the answer is 75/4 =18.75

Answer 4

x=18.75 y=30

restriction problem

A=4xy
300=8x+5y

Answer 5

18.75 ft?