Lacy has $600 set aside to build a rectangular exercise kennel for her dogs. She will buy fencing material for $15/ft. Because the side of an existing barn will be used for one of the sides of the kennel, only three sides need to be fenced.
A.) Determine the dimensions of the kennel that will enclose the maximum area.
B.) What area will be enclosed?
let the width against the barn be x ft
let the length parallel to the barn be y ft
so total length = 600/15 = 40
so she needs to fence 40 ft
2x + y = 40
y = 40-2x
area =xy
= x(40-2x)
= 40x - 2x^2
are you taking Calculus?
if so, d(area)/dx = 40 - 4x = 0 for a max of area
x = 10
then y = 20
so max area = xy = 200 ft^2, when the width is 10 ft and the length is 20 ft
if not Calculus
area = 40 - 2x^2 is a parabola opening down.
we need its vertex:
the x of the vertex is -b/(2a) = -40/-4 = 10
sub in :
area = 40(10) - 2(10^2) = 200 t^2
same conclusion
My question is very similar to this one, except they say that Lacy needs the area of the kennel to be 150ft^2. So what I did was take the known information above, where she can fence 40 feet. The I figured out what times what would equal 150 and add to get 40, (15*10) So since it is along the barn wall, you can have the longer sides to equal 15 and the barn wall side to equal 10. If you add the three sides up, it will equal 40. ;)
Hope this helps!
YOU GALS SO HELPFUL I GOT A 100% THANKS SOOOOO MUCH ('......')
To determine the dimensions of the kennel that will enclose the maximum area, we need to use optimization. We know that the kennel has only three sides that need to be fenced, as one side will be the existing barn. Let's assume the width of the kennel is 'x' and the length is 'y'.
A.) To maximize the area, we need to write an equation for the area of the kennel and then maximize it.
The area of a rectangle is given by the formula: Area = length * width.
In this case, the area of the kennel will be: Area = x * y.
Now, let's write an equation for the perimeter of the kennel. Since there are only three sides that need fencing, the total length of the fencing required will be: Perimeter = x + 2y.
However, we need to consider the cost of the fencing material. The cost of material will be: Cost = $15 * Perimeter.
Given that Lacy has $600 set aside, we can set up the equation: $600 = $15 * Perimeter.
Now, let's solve for Perimeter in terms of x and y:
$600 = $15 * (x + 2y)
Divide both sides by $15:
40 = x + 2y
We can rearrange this equation to get: x = 40 - 2y.
Now, substitute the value of x in the area equation:
Area = (40 - 2y) * y
Area = 40y - 2y^2
To find the dimensions of the kennel that enclose the maximum area, we need to find the value of y that maximizes the Area equation.
B.) To find the area that will be enclosed, we need to substitute the value of y (that maximizes the area) back into the Area equation: Area = 40y - 2y^2.
To find the maximum area, we can either take the derivative of the Area equation with respect to y and set it to zero, then solve for y, or we can graph the Area equation and find the maximum point on the graph.
Once we have the value(s) of y, we can substitute it back into the equation x = 40 - 2y to find the corresponding value(s) of x.
This will give us the dimensions of the kennel that will enclose the maximum area, as well as the maximum area itself.