• May 26th 2011, 01:05 PM
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.
• May 26th 2011, 01:09 PM
TheEmptySet
Quote:

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

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

\$\displaystyle A=lw\$

Now you can use the first equation to eliminate a variable from the 2nd then find the maximum of the resulting quadratic.
• May 26th 2011, 01:11 PM
pickslides
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.
• May 26th 2011, 01:46 PM
bkbowser
Ah so it should be something like x(50-2x).
• May 26th 2011, 03:15 PM
pickslides
Yep A =x(50-2x)

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

Hint: Half way between the zeros.