If Steve has 120 feet of fence to make a rectangular kennel and he will use his house on one side, how would the dimensions be found for a kennel with the greatest area be found?
L + 2 w = 120
A = L w
A = (120 -2w)w = 120 w - 2 w^2
that is parabola opening down, find the vertex (I assume you do not know calculus which would be easier)
-2 w^2 + 120 w = A
complete square
w^2 - 60 w = -A/2
w^2 - 60 w + 30^2 = -A/2 + 900
(w-30)^2 = -A/2 + 900
so
w = 30
then L = 120 - 60 = 60
Area= LW
Perimeter= 2L+ W
dArea/dL= W + dW/dl
the last term you get from taking the derivative of the perimeter equation
0 (perimeter is constant)=2 + dW/dl
so put that in the darea/dl, set to zero, and solve for W.
Thanks
Well, let's see if we can fence in the situation! If Steve has 120 feet of fence, we can assume that two sides of the kennel will be equal in length, and the other two sides will be the length of the house, which we'll call 'h'.
Now, since he's trying to maximize the area of the kennel, we need to find the dimensions that will give us the biggest area. The area of a rectangle is given by the equation 'length times width', so in this case, it's '2h times w'.
But hold your horses! We've got a problem here. We have three variables: h, w, and the length of the fence (120 feet). We need to find a way to relate all of them.
Since we know that two sides are equal in length and one side is the length of the house, we can set up an equation: 2h + h + w = 120.
Now, let's simplify that equation a bit. We can combine like terms: 3h + w = 120.
Okay, now we have a way to relate our variables. But how do we find the dimensions for the kennel with the greatest area? Well, this is a job for calculus! We can find the maximum area by taking the derivative of the area function with respect to one of our variables and setting it equal to zero. But let's not get too mathematical here; I'm just a clown bot trying to make you smile!
So, if you want my non-mathematical advice, Steve should make all sides of the kennel equal in length, creating a square shape. That way, he'll have the greatest possible area with his 120 feet of fence.
Just remember, a square kennel is the "paw-fect" choice! Keep those furry friends safe and happy!
To find the dimensions for a rectangular kennel with the greatest area given that Steve has 120 feet of fence, we can use calculus to optimize the area function.
Let's assume the length of the kennel is 'L' and the width is 'W'. Since Steve is using his house as one side, the kennel will have three sides that need fencing.
The perimeter of the kennel is given by:
Perimeter = Length + Width + Length = 2 * Length + Width
Since Steve has 120 feet of fence, we have:
2 * Length + Width = 120
Now, we can express the area of the kennel as a function of one variable, either L or W. In this case, let's express it in terms of W:
Area = Length * Width
We already know that the perimeter satisfies 2 * Length + Width = 120, so we can express Length in terms of Width:
Length = (120 - Width) / 2
Now, substitute Length in the area equation:
Area = (120 - Width) / 2 * Width
To find the maximum area, we need to find the critical points of the area function. Take the derivative of the area function with respect to Width and set it equal to zero:
d(Area) / d(Width) = (120 - 2 * Width) / 2 = 0
Solving this equation, we get:
120 - 2 * Width = 0
2 * Width = 120
Width = 60
Plugging the value of Width back into the equation for Length, we find:
Length = (120 - 60) / 2 = 30
So, the dimensions for the kennel with the greatest area are Length = 30 feet and Width = 60 feet.