Question 438013: Carl wants to construct three rectangular dog-training arenas side-by-side using a total of 340 feet of fencing. What should the overall length and width be in order to maximize the area of the three combined arenas? (let x represent the width).
I used the formula A=3xy to find the area and I came up with:
A=3x(340-4x)/6
I also used the formula P=4x+6y to find the perimeter and I came up with:
Y=(340-4x)/6
I don't know where to go from here.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You are over-thinking the setup here.
Imagine 3 rectangles stacked on top of one another creating another, larger rectangle.
Let the width of this larger rectangle be represented by  . Let the length of the larger rectangle (the sum of the widths of the three small ones) be represented by 
Now the perimeter of this larger rectangle is just  like any other rectangle, but remember that this is three rectangular areas so there are two more pieces of fence each of which measures , so we can write:
to show the relationship between the length and width of the large rectangle and the amount of available fencing.
Next, the area of the large rectangle, which is the area we are asked to maximize is given by:
\ =\ xy)
But if we solve the fencing relationship for we can create a function for the area in terms of the width:
Then, by substitution:
\ =\ -2x^2\ +\ 170x)
Which is a quadratic function of the form:
\ =\ ax^2\ +\ bx\ + c)
where
 ,  , and 
Since this is a quadratic with a negative lead coefficient, the graph is a parabola that opens downward, hence the vertex represents the maximum value of the function.
Use the formula for the -coordinate of the vertex of a parabola to find the overall width that maximizes the area:
}\ =\ 42.5\ \ ) feet
Substitute this value into the equation for the fencing:
\ =\ 340)
and solve for the overall length:
 feet
John
My calculator said it, I believe it, that settles it
|
|
|
