Fencing area

Question

The owner of the Rancho Los Feliz has 3120 yd of fencing to enclose a rectangular piece of grazing land along the straight portion of a river and then subdivide it by means of a fence running parallel to the sides. No fencing is required along the river. (See the figure below.) What are the dimensions of the largest area that can be enclosed?
x = ___ yd
y = ___ yd

What is this area?___ yd2

Equation for fence is 3x+y=3120

Answer 1

3120 yards of fencing

there is one length and three widths with one width dividing the rectangle in the middle

no fencing is needed because of the river

P=3x+y

3120=3x+y

y=3120-3x

A=xy

A=x(3120-3x)

A=3120x-3x²

this is a parabola opening downward so there is a maximum point

the vertex is (h,k) where h=-b/2a

-b/2a=-3120/-6

-b/2a=520 yards

the largest area is always a square, so the dimensions would ordinarily be 520 by 1040 with a fence in the middle

this is really two squares 520 by 520 next to each other but you don't need fencing near the river

therefore you take the extra 520 of fencing and add it to the length of 1040 to get 1560

you have a rectangle with a length of 1560 yards and two widths of 520 yards each (along with a fence of 520 yards down the middle)

the length is 1560 yards and each of the three parallel widths is 520 yards

this gives a total of 1560+520+520+520=3120 yards

length=1560 yards

width=520 yards

The area is A=l*w

A=1560*520

A=811,200 square yards

 

Answer 2

Optimization problem. Let y be side parallel to river. Let x be the length of each of the three sides perpendicular to the river And parallel to each other. Then perimeter 3120=3x+y, or y = 3120-3x. Area = xy = x(3120-3x). Take the derivative, set to 0 and get x=520 yards. Y would then be 1560 yards. Area would be 811,200 square yards.