The perimeter of the fence is a rectangle, of length x and width y, so we have 2x + 2y = 1500. The area is A = xy.
How can you put the area of the corral in terms of one variable? How do you go about finding a maximum area?
-Dan
The perimeter of the fence is a rectangle, of length x and width y, so we have 2x + 2y = 1500. The area is A = xy.
How can you put the area of the corral in terms of one variable? How do you go about finding a maximum area?
-Dan
it's 3 adjacent corrals though (imgur: the simple image sharer). Don't I have to somehow account for the fence that goes between the corrals?
There are two long fences. Calling each length "x", their lengths total 2x. There are four fences connecting them, two at the ends, two separating the three pens. If we call the length of each "y", their length total 4y. The total fencing is 2x+ 4y= 1500 feet and the area is xy square feet. You want to maximize xy subject to the constraint 2x+ 4y= 1500 feet.
One way to do that is to solve 2x+ 4y= 1500 for x= 750- 2y and write the area as . You can find the maximum by completing the square.