SOLUTION: A rancher has 216 feet of fencing to enclose two adjacent rectangular corrals. What dimensions will produce the largest total area? Your answer is: What is the maximum total area

Algebra ->  Rectangles -> SOLUTION: A rancher has 216 feet of fencing to enclose two adjacent rectangular corrals. What dimensions will produce the largest total area? Your answer is: What is the maximum total area      Log On


   

Question 1033564: A rancher has 216 feet of fencing to enclose two adjacent rectangular corrals. What dimensions will produce the largest total area?
Your answer is:
What is the maximum total area?
Your answer is:
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
The shape to maximize the rectangular fenced area is a square. One of the sides happens three times because the rancher wants to make two adjascent area regions.

The dimensions can be x and y, for the entire area.
Total fencing, 2x+y+x=216, just picking x for the barrier piece of fence between the two adjascent regions. Let A be area, so A=xy. Try drawing a figure to help this all make sense.

system%28A=xy%2C3x%2B2y=216%29.

Do you know what to do from here?

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
A rancher has 216 feet of fencing to enclose two adjacent rectangular corrals. What dimensions will produce the largest total area?
Your answer is:
What is the maximum total area?
Your answer is:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

I will assume that we have two adjacent rectangular corrals with the same dimensions L (length) and W (width), 
One common side (the length) is common for two corrals and the fence is installed along all sides, 
including the common interior side, although it is not stated directly in the condition.

If so, then we have this equation

3L + 4W = 216, 

and we need to maximize the value f = L*2W.

Express 2W via L from (1): 2W = %28216-3L%29%2F2, and substitute it into L*2W. You will get 

f = L%2A%28216-3L%29%2F2 = %281%2F2%29%2AL%2A%28216-3L%29.

To maximize %281%2F2%29%2AL%2A%28216-L%29 means the same as to maximize L%2A%28216-3L%29. So, we can take off this multiplier 1%2F2.

Now we need to find the value of L which maximize this quadratic function

g(L) = L*(216-3L) = -3L%5E2+%2B+216L.

Those who know calculus can easily get the answer: L = 216%2F6 = 36 feet.

The others can determine the vertex of the parabola g(L) = -3L%5E2+%2B+216L using formulas from algebra.

L = -b%2F2a = -216%2F%28-6%29 = 216%2F6 = 36 feet and get the same answer.

Answer. The length of the corral should be 36 feet, the width of each corral is %28216-3%2A36%29%2F4 = %28216-108%29%2F4 = 27 feet.

        The total area of the two corrals is 2*36*27 = 1944 square feet.