# Thread:

1. ## Presumabley a Quadratic word problem

A rectangular field adjacent to a river must be fenced on 3 sides but not on the river bank. What is the largest area that can be enclosed if 50 yards of fencing is used?

The chapter is on Quadratic Functions; Graphing them using intercepts and turning points specifically. Given this I assume that the question wants me to place it into a x^2 format.

First off I can't figure out how to set it up as a quadratic. Secondly the best result I can get is when the sides are equal, or 50/3, which is approximately equal to 277. The answer is 312.

2. Originally Posted by bkbowser
A rectangular field adjacent to a river must be fenced on 3 sides but not on the river bank. What is the largest area that can be enclosed if 50 yards of fencing is used?

The chapter is on Quadratic Functions; Graphing them using intercepts and turning points specifically. Given this I assume that the question wants me to place it into a x^2 format.

First off I can't figure out how to set it up as a quadratic. Secondly the best result I can get is when the sides are equal, or 50/3, which is approximately equal to 277. The answer is 312.
Since the pen is next to a river we only need fence on 3 sides.

Let the side parallel to the river be l and the side perpendicular to the river be w.

Then using the area and perimeter formula for a rectangle we get

$P=l+2w \iff 50=l+2w$

$A=lw$

Now you can use the first equation to eliminate a variable from the 2nd then find the maximum of the resulting quadratic.

3. Start by saying the 50 yards is equal to the sum of the two sides perpendicular (x) and the side that is parallel to the river (y)

You get x+x+y=50 or 2x+y = 50

Now the area is just lenth times width or x times y. So in the above equation solve for y, so area can be a function of just x.

4. Ah so it should be something like x(50-2x).

5. Yep A =x(50-2x)

Now where has this value a maximum? Or where is the turning point

Hint: Half way between the zeros.
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