i am working with completing the square in parabolas and there's this word problem i just cannot solve..a farmer wants to make a rectangular corral along the side of a large barn and has enough materials for 60m of fencing. Only 3 sides must be fenced, since the barn wall will form a 4th side. What width of rectangle should the farmer use so that the maximum area is enclosed?.. please help!:(
Assume the corral will not be as long as the barn, and that is goes a distance x away from the barn. There are thus 2 sides with length x and one side with length 60-x. the total area enlosed is
A = x(60-2x)
Choose the value of x that maximizes A by completing the square.
A = -2x^2 + 60 x
= -2(x^2 -30x + 225) + 225
= ??? You finish it. That is a perfect square within the ()
To solve this problem, we need to find the width of the rectangle that will result in the maximum area.
Let's start by drawing a diagram to visualize the problem. We have a rectangle with width "w" and length "L" (which is equal to the height of the barn). The barn wall acts as the fourth side of the rectangle.
Since we need to fence three sides of the rectangle, we use the formula for the perimeter of a rectangle:
Perimeter = 2 * (width + length)
In this case, the perimeter is given as 60m, so we can set up the equation as follows:
60 = 2 * (w + L)
Simplifying this equation gives us:
30 = w + L
Since the length of the rectangle is equal to the height of the barn, we can rewrite the equation as:
30 = w + h
Now, we need to express the area of the rectangle in terms of the width (w). The area of a rectangle is given by the formula:
Area = width * length
Substituting "h" for the length, we have:
Area = w * h
Now let's express the height in terms of the width using the perimeter equation we found earlier.
30 = w + h
Rearranging the equation gives us:
h = 30 - w
Substituting this expression into the area equation, we have:
Area = w * (30 - w)
Now we have the formula for the Area in terms of the width (w).
To find the width that maximizes the area, we need to take the derivative of the area equation with respect to w, set it to zero, and solve for w. This will give us the critical points that can correspond to a maximum area.
d(Area)/dw = 30 - 2w
Setting this derivative equal to zero, we have:
30 - 2w = 0
Solving for w, we get:
w = 15
So, the width of the rectangle that will result in the maximum area is 15 meters.
To find the corresponding length (h), we can substitute this value of w back into the perimeter equation:
30 = 15 + h
Solving for h, we get:
h = 15
Therefore, the rectangle with a width of 15 meters and a length (height) of 15 meters will enclose the maximum area.